Contents / Algebra / Exponents, Radicals & Logarithms

Chapter 7

Exponents, Radicals & Logarithms

Exponent rules, roots, rational exponents, exponential functions, logarithms, and radical equations.

Introduction

Exponent rules, roots, rational exponents, exponential functions, logarithms, and radical equations.

Exponentiation is the operation of raising a base to an exponent (power). The notation a^n means a multiplied by itself n times when n is a positive integer.

It is one of the most important operations in mathematics because it appears in growth models, scientific notation, polynomials, and almost every branch of higher math. The basic rules of exponents let us simplify expressions without expanding everything.

7.1Definition for Positive Integers

Definition 7.1 (Exponentiation). For a positive integer $n$ ,

a^{n} = n factors a \cdot a \dots a,

and in particular $a^{1} = a$ .

Bases can be any real (or complex) number; the exponent tells how many times to multiply the base by itself.

Intuition. Exponent is a shorthand for repeated multiplication, just like multiplication is shorthand for repeated addition.

2^5 is 2 used as a factor five times: 22222 = 32.

Example 7.2. Compute $3^{4} = 3 \times 3 \times 3 \times 3 = 81$ .

Solution.

Compute $(- 2)^{3} = (- 2) \times (- 2) \times (- 2) = - 8$ . (Odd power keeps the negative sign.)

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Example 7.3. Compute $2^{4} \times 3^{2}$ (two different bases, so handle each separately).

Solution.

$2^{4} = 2 \times 2 \times 2 \times 2 = 16$ and $3^{2} = 3 \times 3 = 9$ .
Multiply the results: $16 \times 9 = 144$ .
Note you cannot add the exponents here: the bases differ, so $2^{4} \times 3^{2} \neq = 6^{anything}$ in general.

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Example 7.4. Compute $(1/2)^{2} = (1/2) \times (1/2) = 1/4$ . A fraction between $0$ and $1$ raised to a power gets smaller, which matches $1/4 < 1/2$ .

Exercises for §7.1 →

7.2Zero and Negative Exponents

Definition 7.5 (Zero and negative exponents). For $a \neq = 0$ ,

a^{0} = 1, a^{- n} = \frac{1}{a ^{n}} .

The first is a convention chosen to preserve the product rule, since $a^{m} \cdot a^{0} = a^{m}$ .

These definitions make the exponent rules work uniformly for all integers.

Negative exponents move the factor to the denominator (or numerator if already in denominator) and flip the sign of the exponent.

Intuition. Anything (nonzero) to the zero power is "one copy of the empty product," which is defined as 1.

Negative power means "one over" the positive power; it is a reciprocal.

Example 7.6. $5^{0} = 1$ , $(x y)^{0} = 1$ (for $x y \neq = 0$ ).

Solution.

$7^{- 1} = 1/7$ , $x^{- 2} = 1/ x^{2}$ .

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Example 7.7.

(2/3)^{- 2} = (3/2)^{2} = 9/4

Exercises for §7.2 →

7.3Properties and Order of Operations

Theorem 7.8 (Laws of exponents). For appropriate bases,

a^{m} \cdot a^{n} = a^{m + n}, (a^{m})^{n} = a^{mn}, (ab)^{n} = a^{n} b^{n},

\frac{a ^{m}}{a ^{n}} = a^{m - n} = \frac{1}{a ^{n - m}}, (\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}} .

When evaluating, follow order of operations: parentheses first, then exponents (right-to-left for a^{b^c}), then * and / left to right, then + -.

Intuition. When multiplying same base, add the exponents (like combining lengths of repeated groups).

Power of a power: multiply the exponents (layers of repetition).

Everything distributes over multiplication in the base.

Example 7.9.

x^{3} \cdot x^{5} = x^{3 + 5} = x^{8}

Solution.

$(y^{2})^{4} = y^{2 \cdot 4} = y^{8}$ .
$(2 x)^{3} = 2^{3} x^{3} = 8 x^{3}$ .

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Example 7.10.

(3^{2} \cdot 3^{3}) / 3^{4} = 3^{2 + 3} / 3^{4} = 3^{5} / 3^{4} = 3^{5 - 4} = 3^{1} = 3

Example 7.11. Evaluate the stacked exponent $2^{3^{2}}$ . Exponentiation is right-associative, so you resolve the top exponent first: $2^{(3^{2})}$ , not $(2^{3})^{2}$ .

Solution.

Top first: $3^{2} = 9$ .
Then $2^{9} = 512$ .
Contrast: reading left-to-right would give the wrong $(2^{3})^{2} = 8^{2} = 64$ . The right-associative reading $512$ is the correct one.

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Exercises for §7.3 →

Common Mistakes

Treating a^n as a * n: Exponent is repeated multiplication, not scaling by the exponent.
Adding exponents on multiplication of different bases: 2^3 * 3^3 = (2*3)^3 = 6^3, not 2^3 * 3^3 by adding.
Forgetting a^0 = 1 for nonzero a: 100^0 is 1, not 0 or 100.
Sign errors with negative bases and even/odd powers: (-2)^4 is positive, (-2)^3 is negative.
Misapplying the power rule to sums: (a + b)^2 ≠ a^2 + b^2 (the binomial theorem gives the cross term).
Negative exponent on the whole fraction without flipping: (a/b)^{-1} = b/a , not 1/(a/b) incorrectly handled.
Evaluating left to right for stacked exponents: 2^{3^2} = 2^9 = 512, not (2^3)^2 = 64. Right-associative.

The power rules (or laws of exponents) are the algebraic shortcuts that let us manipulate expressions containing exponents without expanding them into long products.

The two core ideas are the product rule and the power-of-a-power rule; from them the others follow. Mastering these rules is essential for simplifying algebraic expressions, solving exponential equations, and working with polynomials and scientific notation.

7.4Product Rule

When multiplying powers that have the identical base, keep the base and add the exponents:

a^{m} \cdot a^{n} = a^{m + n} .

Proof. This works because you are simply counting the total number of times the base is used as a factor.∎

Intuition. Same base? Add the "how many times" numbers.

x^2 * x^3 is x used 2 times then 3 more times = x used 5 times total.

Example 7.12.

x^{5} \cdot x^{3} = x^{5 + 3} = x^{8}

Solution.

$y \cdot y^{4} \cdot y^{2} = y^{1 + 4 + 2} = y^{7}$ .

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Example 7.13. $(2^{3}) \cdot (2^{2}) = 2^{3 + 2} = 2^{5} = 32$ . (8 * 4 = 32, same result.)

Exercises for §7.4 →

7.5Power of a Power and Power of a Product

Raising a power to another power multiplies the exponents, and raising a product to a power raises each factor to that power:

(a^{m})^{n} = a^{mn}, (ab)^{n} = a^{n} b^{n} .

These rules let you "peel" exponents inward or outward.

Intuition. A power of a power is like repeating the repetition: multiply the counts.

Power on a product "distributes" the exponent to each piece inside the parentheses.

Example 7.14.

(x^{3})^{4} = x^{3 \cdot 4} = x^{12}

Solution.

$(3^{2})^{5} = 3^{2 \cdot 5} = 3^{10}$ .

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Example 7.15.

(2 x y^{2})^{3} = 2^{3} \cdot x^{3} \cdot (y^{2})^{3} = 8 x^{3} y^{6}

Exercises for §7.5 →

7.6Quotient Rule and Combined Use

Dividing powers of the same base subtracts the exponents, and a power on a fraction goes to numerator and denominator separately:

\frac{a ^{m}}{a ^{n}} = a^{m - n} (a \neq = 0), (\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}} .

When the rules are combined with negative and zero exponents, almost any expression with integer powers can be simplified to a single power or a fraction of powers with positive exponents.

Intuition. Dividing same base: subtract the exponents (cancel the overlapping factors).

Power on a fraction distributes to num and den separately.

Example 7.16.

x^{7} / x^{2} = x^{7 - 2} = x^{5}

Solution.

$(x^{4} / y^{3})^{2} = x^{4 \cdot 2} / y^{3 \cdot 2} = x^{8} / y^{6}$ .

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Example 7.17. $(a^{3} b^{- 2})^{2} / (a^{- 1} b^{4}) = (a^{6} / b^{4}) / (1/ (a b^{4}))$ . First simplify: numerator $a^{6} b^{- 4}$ , divide by $a^{- 1} b^{4} = a^{6} b^{- 4} \cdot a^{1} / b^{4} = a^{7} b^{- 8} = a^{7} / b^{8}$ .

Exercises for §7.6 →

Common Mistakes

Adding exponents when bases differ: 2^3 * 3^2 is not 2^5 or 6^something; compute separately or factor.
Multiplying exponents on product of powers with same exponent: a^2 * b^2 = (a b)^2 , not a^4 b^something unless using other rules.
Forgetting parentheses when bases have coefficients: 2x^3 squared is (2x)^3? No, (2 x^3 )^2 vs 2 (x^3)^2.
Subtracting exponents on multiplication: Product adds, quotient subtracts.
Applying rules to sums or differences: (a + b)^n does not equal a^n + b^n in general.
Ignoring the condition a ≠ 0 for negative or zero exponents: 0^0 and negative powers of zero are undefined.
Wrong association on stacked powers: 2^{3^2} = 2^9 not 8^2 (exponentiation is right-associative).

The (principal) square root of a non-negative real number x, written √x or x^{1/2}, is the unique non-negative real number y such that y^2 = x.

Square roots are the inverse operation to squaring. They appear in the quadratic formula, distance formula, standard deviation, and countless geometric and physical formulas. Understanding the domain (x ≥ 0), the principal root convention, and how to simplify and solve with them is fundamental.

7.7Definition and the Principal Root

Definition 7.18 (Principal square root). For $x \geq 0$ , the principal square root $x$ is the unique $y \geq 0$ with

y^{2} = x .

Every positive $x$ has two real square roots, $+ y$ and $- y$ , but the symbol $x$ always returns the non-negative one. The negative root is written $- x$ .

√0 = 0. Negative numbers have no real square root (they have imaginary roots involving i).

Intuition. Squaring forgets the sign; the square root symbol "chooses" the positive answer to make functions well-defined.

√9 = 3, not -3, even though both square to 9.

Example 7.19. $25 = 5$ (the principal). The equation $x^{2} = 25$ has solutions $\pm 5$ .

Example 7.20.

1/4 = 1/2

Exercises for §7.7 →

7.8Properties and Simplifying Square Roots

Theorem 7.21 (Properties of square roots). For $a, b \geq 0$ ,

ab = a b, \frac{a}{b} = \frac{a}{b} (b > 0), a^{2} = ∣ a ∣.

The absolute value appears in the last one because the principal root is non-negative.

To simplify √N where N is not a perfect square, factor out the largest perfect square factor.

Intuition. Roots distribute over multiplication and division (like exponents).

But √(a^2) "folds" the negative back to positive, hence absolute value.

Example 7.22.

50 = 25 \cdot 2 = 52

Solution.

$12 x^{3}$ for $x \geq 0 = 4 \cdot 3 x^{2} \cdot x = 2∣ x ∣ 3 x = 2 x 3 x$ .

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Example 7.23.

18 + 8 = 32 + 22 = 52

Exercises for §7.8 →

7.9Operations and Solving Equations

When multiplying or adding roots, simplify first, then combine like terms.

Method 7.24 (Solving an equation with a square root).

Isolate the root on one side.
Square both sides.
Solve the resulting equation.
Check every candidate in the original equation — squaring can introduce extraneous solutions.

The domain restrictions must be respected throughout.

Intuition. Simplifying first makes adding or multiplying easier (like reducing fractions before operating).

Squaring to solve is powerful but "lies" sometimes; always verify.

Example 7.25. Solve $3 x + 1 = 5$ .

Solution.

Square: $3 x + 1 = 25 \to 3 x = 24 \to x = 8$ .
Check: $24 + 1 = 25 = 5$ . Good.

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Example 7.26. Solve $x - 2 = x - 4$ (note right side must be $\geq 0$ so $x \geq 4$ ).

Solution.

Square: $x - 2 = (x - 4)^{2} = x^{2} - 8 x + 16$ .
$0 = x^{2} - 9 x + 18 = (x - 3) (x - 6)$ .
Candidates $x = 3, 6$ . Only $x = 6$ satisfies original ( $x = 3$ gives $1 = - 1$ which is false and right side negative). $x = 6$ : $4 = 2$ , $6 - 4 = 2$ . Good.

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Exercises for §7.9 →

Common Mistakes

Writing √(a + b) = √a + √b: False in general (e.g. √(4+9)=√13 ≠ 2+3).
Forgetting the absolute value in √(x^2) = |x|: For negative x the principal root is positive.
Ignoring domain when simplifying or solving: √(x^2 - 1) requires |x| ≥ 1.
Not checking solutions after squaring: Extraneous roots appear when both sides were not non-negative or when squaring changed the equality set.
Confusing the symbol √x with "both roots": The symbol is the principal (≥0) root only; use ± when solving x^2 = k.
Dividing by zero or taking roots of negative in real context: Always state restrictions.
Simplifying √(x^2 + 2x + 1) as x + 1 without | |: It is |x + 1|.

The nth root of a number x, written √[n]x or x^{1/n}, is a number y such that y^n = x.

When n=2 it is the square root; n=3 the cube root, etc. Odd roots are defined for all real x and return one real value (positive or negative). Even roots are defined only for x ≥ 0 and return the non-negative principal root.

nth roots generalize square roots and appear in solving polynomial equations, in radical expressions, and when working with rational exponents.

7.10Definition and Even vs Odd Roots

Definition 7.27 (nth root). For an integer $n \geq 2$ , an $n$ th root of $x$ is a number $y$ with $y^{n} = x$ .

If $n$ is odd: for any real $x$ there is exactly one real $y$ with $y^{n} = x$ , and $y$ can be negative if $x$ is.
If $n$ is even: a real $y$ exists only for $x \geq 0$ , and we take the principal root $y \geq 0$ .

The symbol $n x$ always denotes the principal (non-negative when it makes sense) root.

Intuition. Odd roots (cube, 5th, ...) are like "undo" for any sign; they preserve sign information.

Even roots are like square roots: they "forget" sign and only accept non-negative inputs in reals.

Example 7.28. $3 - 8 = - 2$ because $(- 2)^{3} = - 8$ .

Solution.

$481 = 3$ (principal), even though $(- 3)^{4} = 81$ also.

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Example 7.29.

532 = 2

Exercises for §7.10 →

7.11Properties of nth Roots

Theorem 7.30 (Properties of nth roots). Whenever the expressions are defined in the reals,

n ab = n a n b, n \frac{a}{b} = \frac{n a}{n b} (b \neq = 0),

n a^{n} = {∣ a ∣, a, n even, n odd .

These mirror the square root properties but with care for the even/odd distinction on signs and domains.

Intuition. Roots "distribute" over multiplication and division when everything stays real.

For even roots the absolute value appears on √n because the root returns positive.

Example 7.31.

3 8 \cdot 27 = 38 \cdot 327 = 2 \cdot 3 = 6

Solution.

$4 16 x^{4} = 416 \cdot 4 (x^{4}) = 2∣ x ∣$ (for even root).

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Example 7.32. $3 x^{3} /8 = x /2$ (for all real $x$ ).

Exercises for §7.11 →

7.12Simplifying and Solving with nth Roots

To simplify, factor the radicand to pull out perfect nth powers.

When solving equations with nth roots, raise both sides to the nth power. For odd n this is one-to-one and rarely introduces extraneous solutions. For even n (like square roots) always check candidates.

Intuition. Pull out "perfect groups" of n identical factors, leaving the leftover under the root.

Raising to n "undoes" the root, but even n can create sign ambiguity or extras.

Example 7.33. Simplify $3 54 x^{4} = 3 27 \cdot 2 x^{3} \cdot x = 3 x 3 2 x$ .

Example 7.34. Solve $4 x - 1 = 2$ .

Solution.

Raise to 4th: $x - 1 = 16$ , $x = 17$ .
Check: $416 = 2$ (principal), good. (Note $x - 1$ must be $\geq 0$ .)

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Exercises for §7.12 →

Common Mistakes

Taking even roots of negatives in real numbers: Not defined (use complex if needed).
Writing √[n](a + b) = √[n]a + √[n]b: Almost never true.
Forgetting | | for even roots of powers: √4 = |x|, not x.
Not checking after raising to even power: Extraneous solutions common.
Confusing index n with the exponent on the variable: ∛(x^3) = x, but √(x^2) = |x|.
Assuming all roots have two real solutions: Only even powers in the equation typically do; odd root equations have one.
Domain slips when simplifying products inside even roots: If a or b negative the product may still be positive, but individual roots may not be real.

A rational exponent is an exponent that is a fraction p/q in lowest terms. By definition, for a ≥ 0 (or appropriate domain),

a^{p/q} = (a^{1/q})^p = (a^p)^{1/q} .

Rational exponents provide a uniform way to express roots and powers together and extend the familiar integer exponent rules to fractional cases.

They are the bridge between radical notation and exponential notation and are heavily used in calculus, algebra, and modeling.

7.13Definition as Roots and Powers

Definition 7.35 (Rational exponent). The denominator of the exponent indicates the root and the numerator indicates the power:

a^{m / n} = (a^{1/ n})^{m} = (a^{m})^{1/ n} .

That is, take the $n$ th root of $a$ and then raise to the power $m$ — or raise first and then take the root; both agree wherever they are defined.

For negative bases, the expression is real when the root index (denominator in reduced form) is odd.

Intuition. The fraction tells you "root first or power first" — two ways to the same place when it exists.

8^{2/3} is the cube root of 8, then squared (or 8 squared then cube root) = 4.

Example 7.36.

2 7^{2/3} = (2 7^{1/3})^{2} = 3^{2} = 9

Example 7.37. $(- 32)^{3/5}$ : 5th root of $- 32$ is $- 2$ , then $^{3} = - 8$ . (Denom 5 odd.)

Exercises for §7.13 →

7.14Properties of Rational Exponents

Theorem 7.38 (The exponent laws extend to rational exponents). Whenever the expressions are defined in the reals, all the integer power rules hold for rational $r$ and $s$ :

a^{r} \cdot a^{s} = a^{r + s}, (a^{r})^{s} = a^{rs}, (ab)^{r} = a^{r} b^{r}, \frac{a ^{r}}{a ^{s}} = a^{r - s} .

This unifies roots and powers under one set of rules.

Intuition. You can treat fractional exponents almost like integers for adding, multiplying, etc., as long as you don't step outside the domain.

The arithmetic of the fractions does the combining of roots and powers.

Example 7.39.

x^{1/2} \cdot x^{3/4} = x^{1/2 + 3/4} = x^{5/4}

Solution.

$(x^{2/3})^{3} = x^{2}$ .

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Example 7.40.

(8 x^{6})^{1/3} = 2 x^{2}

Exercises for §7.14 →

7.15Simplifying Expressions and Solving Equations

Rewrite rational exponents as radicals when helpful for simplification, or vice versa.

To solve equations with rational exponents, raise both sides to the reciprocal power (or clear by raising to lcm of denominators), then check solutions against the domain (especially non-negative requirements for even roots).

Intuition. Convert to roots to see what is "inside" the root, or to exponents to use exponent arithmetic.

Raising to the reciprocal "undoes" the fractional power, but domain and extraneous checks still needed.

Example 7.41. Simplify $(x^{3/2} y^{- 1/2})^{4} = x^{6} / y^{2}$ .

Example 7.42. Solve $x^{3/2} = 8$ .

Solution.

Raise both sides to $2/3$ : $x = 8^{2/3} = 4$ . ( $x \geq 0$ required.)
Check: $4^{3/2} = (4^{1/2})^{3} = 2^{3} = 8$ . Good.

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Example 7.43. Solve $x^{4/3} = 16$ for real $x$ .

Solution.

Rewrite the left side as $(x^{1/3})^{4} = 16$ . Since the root index is $3$ (odd), the cube root is defined for negative bases too, so both signs are possible.
Let $y = x^{1/3}$ . Then $y^{4} = 16$ , so $y = \pm 2$ (both real fourth roots of $16$ ).
Recover $x = y^{3}$ : from $y = 2$ , $x = 8$ ; from $y = - 2$ , $x = - 8$ .
Check both: $8^{4/3} = (8^{1/3})^{4} = 2^{4} = 16$ , and $(- 8)^{4/3} = ((- 8)^{1/3})^{4} = (- 2)^{4} = 16$ . Both solutions are valid.

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Exercises for §7.15 →

Common Mistakes

Using even denominator with negative base without checking: (-4)^{1/2} not real; (-8)^{2/3} is real because denom 3 odd.
Adding exponents without common denominator: 1/2 + 1/3 = 5/6, not 2/5 or 1/6.
Assuming (a^{p/q})^r always = a^{p r / q} without domain: For negative a and certain fractions it may leave reals.
Forgetting to convert to common base or root when simplifying mixed radical/exponent forms.
Raising both sides to a power without considering the principal root or sign: Can miss solutions or add extras.
Writing a^{1/2 + 1/3} as a^{1/5}: Wrong arithmetic on the exponent fraction.
Ignoring that 0 to a negative or zero fractional power is undefined.

An exponential function has the form f(x) = a * b^x where a ≠ 0, b > 0, b ≠ 1, and the variable x appears in the exponent.

Exponential functions model phenomena with constant percent growth or decay: populations, compound interest, radioactive decay, cooling, and many natural processes. Their graphs are smooth curves that are always positive, have a horizontal asymptote, and grow (or decay) faster and faster.

7.16Definition and Basic Shape

Definition 7.44 (Exponential function). An exponential function has the form

f (x) = a b^{x}, a \neq = 0, b > 0, b \neq = 1,

with the variable in the exponent. Here $a$ is the initial value, or y-intercept, since $f (0) = a$ ; and $b$ is the growth or decay factor.

The graph is continuous, always above the x-axis, and has the horizontal asymptote $y = 0$ .

For b > 0, b ≠ 1 the function is one-to-one, so it has an inverse (the logarithm).

Intuition. Each time x increases by 1, the output is multiplied by b (not added).

"Percent change is constant" is the key: +50% each step means *1.5, no matter the current size.

Example 7.45. $f (x) = 200 \cdot (1.1)^{x}$ : starts at $200$ when $x = 0$ , grows $10%$ per unit $x$ .

Solution.

At $x = 1$ : $220$ ; $x = 2$ : $242$ , etc.

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Example 7.46. $g (x) = 50 \cdot (0.8)^{x}$ : starts at $50$ , decays $20%$ per unit time, approaches $0$ .

Exercises for §7.16 →

7.17Growth versus Decay and the Role of the Base

Proposition 7.47 (Growth versus decay). Take $a > 0$ . If $b > 1$ the function shows exponential growth: as $x \to + \infty$ , $f (x) \to + \infty$ , and as $x \to - \infty$ , $f (x) \to 0$ .

If $0 < b < 1$ it shows exponential decay: as $x \to + \infty$ , $f (x) \to 0$ , and as $x \to - \infty$ , $f (x) \to + \infty$ .

The base b is often written 1 + r where r is the growth rate (r > -1, r ≠ 0).

Intuition. b >1 is "more than one copy each time step" → explodes upward.

b <1 is "less than one copy" → shrinks toward zero.

Example 7.48. A population of $1000$ grows $5%$ per year, so $P (t) = 1000 \cdot (1.05)^{t}$ . Here $b = 1 + r = 1.05 > 1$ , so this is growth.

Solution.

After $2$ years: $P (2) = 1000 \cdot (1.05)^{2} = 1000 \cdot 1.1025 = 1102.5 \approx 1103$ .
Sanity check: two $5%$ increases should slightly exceed a single $10%$ increase ( $1100$ ) because the second $5%$ is applied to the already-larger amount. Indeed $1102.5 > 1100$ . Good.

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Example 7.49. A drug decays with a half-life of $6$ hours: amount $= A_{0} \cdot (1/2)^{t /6}$ . Here $b = 1/2 < 1$ , so this is decay.

Solution.

Starting from $A_{0} = 80$ mg, after $6$ hours: $80 \cdot (1/2)^{6/6} = 80 \cdot (1/2)^{1} = 40$ mg (exactly one half-life, so half remains).
After $12$ hours: $80 \cdot (1/2)^{12/6} = 80 \cdot (1/2)^{2} = 20$ mg. Each half-life halves the amount, and as $t \to \infty$ the amount $\to 0$ .

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Exercises for §7.17 →

7.18Transformations of Exponential Functions

The usual transformations apply:

f(x) + k : vertical shift.
a f(x) : vertical stretch/compression/reflection (sign of a).
f(x - h) : horizontal shift.
f(c x) : horizontal stretch/compression (c affects the "speed" of growth).

The horizontal asymptote moves with vertical shifts. The y-intercept changes with a and shifts.

Intuition. Think of the bare curve $b^{x}$ as a rubber stamp. Adding $k$ slides the whole stamp (and its "floor," the asymptote) up or down; multiplying by $a$ stretches it taller or flips it upside-down; changing $x$ to $x - h$ slides it sideways.

The one thing that moves the horizontal asymptote is a vertical shift $+ k$ — the floor rises to $y = k$ . Stretching or shifting sideways never lifts the floor.

Example 7.50. Describe $g (x) = 3 \cdot 2^{x - 1} + 4$ as transformations of $f (x) = 2^{x}$ .

Solution.

The $(x - 1)$ shifts right $1$ ; the factor $3$ stretches vertically by $3$ ; the $+ 4$ shifts up $4$ .
The horizontal asymptote moves with the vertical shift only: from $y = 0$ to $y = 4$ .
Check the $y$ -intercept: $g (0) = 3 \cdot 2^{- 1} + 4 = 3/2 + 4 = 11/2$ , which sits above the asymptote $y = 4$ as expected for an upward-opening growth curve.

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Example 7.51. Analyze $g (x) = - 2 \cdot 3^{x - 1} + 5$ : reflect over the $x$ -axis (from the minus sign), stretch vertically by $2$ , shift right $1$ , shift up $5$ .

Solution.

Horizontal asymptote: $y = 5$ (moves with the vertical shift $+ 5$ ).
$y$ -intercept: $g (0) = - 2 \cdot 3^{- 1} + 5 = - 2/3 + 5 = 13/3 \approx 4.33$ .
Sanity check: because the graph is reflected downward (coefficient $- 2$ ), the curve stays below its asymptote $y = 5$ ; indeed $13/3 < 5$ . Good.

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Exercises for §7.18 →

7.19Solving Basic Exponential Equations

Proposition 7.52 (Matching exponents). An exponential function is one-to-one, so for $b > 0$ , $b \neq = 1$ ,

b^{u} = b^{v} ⟺ u = v .

When both sides of an equation can be written as powers of the same base, equate the exponents.

Otherwise take logarithms (see logarithm lesson) or use numerical/graphical methods.

Always consider domain (expression defined) and check solutions.

Intuition. If the bases match, the exponents must match (because exponential is one-to-one).

"Undo" the exponential with a log when bases don't match.

Example 7.53. Solve $4^{x} = 16$ .

Solution.

$4^{x} = 4^{2}$ (since $16 = 4^{2}$ ), so $x = 2$ .

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Example 7.54. Solve $5^{x - 1} = 125$ .

Solution.

$125 = 5^{3}$ , so $x - 1 = 3$ , $x = 4$ .

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Exercises for §7.19 →

Common Mistakes

Confusing exponential with polynomial (x in base vs exponent): 2^x grows much faster than any x^n.
Thinking b=1 or b<0 is allowed for real exponential functions in basic algebra: Base must be positive and not 1 for the standard real exponential.
Forgetting the horizontal asymptote when sketching or transforming: It moves only with vertical shifts.
Solving b^x = c by taking xth root incorrectly: Better to express c as power of b when possible.
Sign errors with negative a: a can be negative, which reflects the graph over x-axis (still positive? No, then it goes negative, but b^x always >0 so sign follows a).
Ignoring that for 0<b<1 the growth/decay labels flip with direction of x.
Writing (a b)^x = a^x * b^x correctly but then mishandling when a is the variable.

A logarithm is the inverse operation of exponentiation. log_b(a) is the exponent to which the base b must be raised to produce a. In symbols: b^{log_b a} = a and log_b (b^c) = c.

Logarithms turn multiplication into addition, allow us to solve for variables in the exponent, measure orders of magnitude, and appear in the definition of many scales (decibels, Richter, pH) and in calculus as the natural log.

7.20Definition and Notation

Definition 7.55 (Logarithm). For $a > 0$ , $b > 0$ and $b \neq = 1$ ,

lo g_{b} (a) = c ⟺ b^{c} = a .

Here $b$ is the base, $a$ is the argument, and $c$ is the logarithm — that is, the exponent.

Two special cases are worth memorising: $lo g_{b} (1) = 0$ and $lo g_{b} (b) = 1$ .

The expression is only defined for positive a.

Intuition. Log asks "what power gets me there?" If 2^? =8, the answer is the log.

It undoes the exponential, like square root undoes squaring.

Example 7.56. $lo g_{2} (16) = 4$ because $2^{4} = 16$ .

Solution.

$lo g_{10} (0.001) = - 3$ because $1 0^{- 3} = 0.001$ .

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Example 7.57. $lo g_{5} (1) = 0$ , $lo g_{5} (5) = 1$ .

Exercises for §7.20 →

7.21Common Bases and Change of Base

Notation (Common and natural logarithms).

The common log $lo g (x)$ usually means $lo g_{10} (x)$ .
The natural log $ln (x)$ means $lo g_{e} (x)$ , where $e \approx 2.718$ is the base of the natural exponential.

Theorem 7.58 (Change of base). Any base can be converted into any other:

lo g_{b} a = \frac{ln a}{ln b} = \frac{lo g _{10} a}{lo g _{10} b} .

Change of base lets us compute any log using calculators that have only ln or log10.

Intuition. Base 10 is convenient for our decimal system and powers of 10.

Base e arises naturally from continuous growth and calculus.

Change of base is like using a universal translator between different "languages" of logs.

Example 7.59.

lo g_{2} (10) \approx ln (10) / ln (2) \approx 2.302585/0.693147 \approx 3.3219

Example 7.60. $lo g_{3} (81) = lo g (81) / lo g (3) = 4$ (exactly, since $3^{4} = 81$ ).

Exercises for §7.21 →

7.22Properties of Logarithms

Theorem 7.61 (Laws of logarithms). For $m, n > 0$ ,

lo g_{b} (mn) = lo g_{b} m + lo g_{b} n,

lo g_{b} (\frac{m}{n}) = lo g_{b} m - lo g_{b} n,

lo g_{b} (m^{k}) = k lo g_{b} m .

These turn products and powers into sums, which is why logs linearize exponential relationships.

Also $lo g_{b} (b) = 1$ , $lo g_{b} (1) = 0$ , and the inverse properties above.

Intuition. Log of a product is sum of logs (like how exponents add on multiplication).

Power comes out front as a multiplier.

Example 7.62. $lo g (100 x^{3} / y) = lo g 100 + 3 lo g x - lo g y = 2 + 3 lo g x - lo g y$ (base 10).

Example 7.63.

2 lo g_{5} 3 + lo g_{5} 4 = lo g_{5} (3^{2}) + lo g_{5} 4 = lo g_{5} (9 \cdot 4) = lo g_{5} 36

Exercises for §7.22 →

7.23Solving Exponential and Logarithmic Equations

Method 7.64 (Solving exponential and logarithmic equations).

Exponential equation: take the log of both sides (any base), or rewrite both sides with the same base and match exponents.
Logarithmic equation: exponentiate both sides — raise the base to each side — or use the laws of logarithms to combine several logs into one first.
Always check the domain: arguments must be positive and bases valid.

Intuition. To "get the variable out of the exponent," take a log.

To get it out of the log, raise the base to the power of both sides.

Example 7.65. Solve $3^{x} = 27$ .

Solution.

Take log base 3: $x = lo g_{3} 27 = 3$ .
Or $3^{x} = 3^{3}$ , $x = 3$ .

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Example 7.66. Solve $lo g_{4} (x + 3) = 2$ .

Solution.

$4^{2} = x + 3 \to 16 = x + 3 \to x = 13$ .
Check: $lo g_{4} (16) = 2$ , yes.

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Example 7.67. Solve $2 ln x - ln 3 = ln 5$ .

Solution.

$ln (x^{2} /3) = ln 5 \to x^{2} /3 = 5 \to x^{2} = 15 \to x = 15$ (positive).

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Exercises for §7.23 →

Common Mistakes

Log of sum is not sum of logs: log(a + b) ≠ log a + log b.
Forgetting domain restrictions: log of negative or zero undefined; solutions that make arg ≤0 are invalid.
Base of log <0 or =1: Not allowed in real numbers.
Confusing log_b a with a / b or b^a: It is the exponent, not division or power swapped.
Dropping the base when using change of base: Must divide logs of same base.
When solving, not checking after exponentiating or taking logs: Extraneous can appear from domain.
Thinking ln means log base 10: ln is specifically natural log base e.

A radical equation contains the variable inside a root (radical) symbol, such as √(x + 2) = 5 or ∛(2x) - 1 = 4.

The standard solution technique is to isolate the radical and raise both sides to the index of the root to "undo" it. Because raising to a power is not a one-to-one operation in the same way for even roots, extraneous solutions can appear and must be checked.

Radical equations arise when solving distances, times, or inverse power relationships.

7.24Isolating the Radical and Raising to a Power

Method 7.68 (Solving a radical equation).

Isolate one radical on one side of the equation.
Raise both sides to the power equal to the root index — square for square roots, cube for cube roots, and so on.
If another radical remains, repeat the isolate-and-raise process.
Solve the resulting polynomial or simpler equation.

For odd roots the process is usually reversible without extras; for even roots extra caution is required.

Intuition. Get the "trapped" variable alone under the root, then "peel off" the root by raising to the matching power.

Like unwrapping layers; each raise removes one root.

Example 7.69. Solve $x - 1 = 3$ .

Solution.

Square both sides: $x - 1 = 9 \to x = 10$ .
Check: $10 - 1 = 3$ . Good.

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Example 7.70. Solve $3 x + 5 = 2$ .

Solution.

Cube both sides: $x + 5 = 8 \to x = 3$ .
Check: $38 = 2$ . Good.

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Exercises for §7.24 →

7.25Extraneous Solutions and Checking

After raising, every candidate must be substituted back into the original equation.

Reasons for extraneous roots:

Squaring (or even powering) can turn a negative "would-be" value into a positive that satisfies the squared equation but not the original (principal root can't be negative).
The isolated radical expression may have been negative, which is impossible for even roots.

Domain: any even root requires its radicand ≥ 0 in the original.

Intuition. Squaring "hides the sign." A number and its negative both square to the same thing.

The check is the honesty step that keeps only answers that were legal all along.

Example 7.71. Solve $x + 2 = x - 4$ .

Solution.

Square: $x + 2 = (x - 4)^{2} = x^{2} - 8 x + 16$ .
$0 = x^{2} - 9 x + 14 = (x - 2) (x - 7)$ .
Candidates: $x = 2$ or $x = 7$ .
Check $x = 2$ : $4 = 2$ , $2 - 4 = - 2$ . $2 \neq = - 2$ . Discard.
Check $x = 7$ : $9 = 3$ , $7 - 4 = 3$ . Equal. Good. Only $x = 7$ .

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Example 7.72. Solve $2 x - 3 + 4 = 1$ .

Solution.

Isolate: $2 x - 3 = - 3$ (impossible for principal square root). No solution. (Squaring would produce extraneous $x = 6$ that fails check.)

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Exercises for §7.25 →

7.26Equations with Multiple Radicals

Method 7.73 (Equations with several radicals).

Move one radical to each side, if that helps.
Square (or raise) to eliminate one of them.
The cross term usually creates a new radical; isolate it and raise again.
Check all resulting candidates thoroughly.

Sometimes substitution (let u = √x or similar) simplifies.

Intuition. Each squaring can create a "new monster radical" from the cross term; keep isolating and squaring until none remain.

Verification is even more important with multiple layers.

Example 7.74. Solve $x + 3 + x = 3$ .

Solution.

Isolate one: $x + 3 = 3 - x$ .
Square: $x + 3 = 9 - 6 x + x$ .
$3 = 9 - 6 x$ .
$6 x = 6$ .
$x = 1$ .
$x = 1$ .
Check: $1 + 3 + 1 = 2 + 1 = 3$ . Good.

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Example 7.75. Solve $x + 7 - x - 5 = 2$ (domain: $x \geq 5$ so both radicands are $\geq 0$ ).

Solution.

Isolate one radical: $x + 7 = 2 + x - 5$ .
Square both sides fully: $x + 7 = 4 + 4 x - 5 + (x - 5)$ .
Simplify: $x + 7 = x - 1 + 4 x - 5$ , so $8 = 4 x - 5$ .
Isolate the new radical: $x - 5 = 2$ .
Square again: $x - 5 = 4$ , so $x = 9$ .
Check: $9 + 7 - 9 - 5 = 16 - 4 = 4 - 2 = 2$ . Good, and $x = 9 \geq 5$ .

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Exercises for §7.26 →

Common Mistakes

Raising before isolating: The cross terms become messy or you square extra junk.
Forgetting to check: Many textbooks problems are designed to have extraneous roots.
Allowing negative under even root after squaring: The candidate may make radicand negative.
Thinking odd-root equations can have extraneous: They usually don't, but domain of other parts of equation may still exclude.
Not squaring the entire side: When isolating one radical, the other side (which may contain another radical or constants) must be squared completely.
Losing solutions when there are two even roots: Each may contribute ± but principal roots are non-neg; the ± comes from the algebra after.
Domain restrictions from original not carried forward: Even after simplification, original even-root domains must hold.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter