Logarithmic expressions

AP Precalculus· AP Precalculus CED — Exponential and Logarithmic Functions· 14 min read

1. Definition and Logarithm-Exponential Equivalence★★☆☆☆⏱ 3 min

A logarithmic expression is any algebraic expression containing one or more logarithms, which are the inverse functions of exponential functions. Logarithms answer the question: what exponent do I raise a given base to, to get the argument value? On the AP exam, you will primarily work with common logarithms (base 10, written $e x t l o g x$ ) and natural logarithms (base $e$ , written $e x t l n x$ ), though any positive base $b \neq = 1$ is valid. Mastery of this topic is required for nearly all subsequent Unit 2 content, so errors here cascade into higher-weight problems.

📘 Definition

Logarithm-Exponential Equivalence

$y = lo g_{b} x ⟺ b^{y} = x$

The entire definition of a logarithm rests on this inverse relationship with exponentiation. It allows free conversion between exponential and logarithmic forms. Domain restrictions require $b > 0, b \neq = 1, x > 0$ : negative arguments or invalid bases produce non-real results.

📐 Worked Example

Evaluate $lo g_{4} (\frac{1}{64})$

1
Let $y = lo g_{4} (\frac{1}{64})$ . By equivalence, this translates to:
2
$4^y = \frac{1}{64}$
3
Rewrite both sides with base 4 to match the logarithm's base:
4
$4^y = (4^3)^{-1} = 4^{-3}$
5
Exponential functions are one-to-one, so exponents are equal, giving $y = - 3$ . Verify by checking $4^{- 3} = \frac{1}{64}$ , which matches the original argument.

Exam tip:

If you are ever unsure of a logarithm value, convert it back to exponential form to check—this takes 10 seconds and eliminates almost all sign errors.

2. Core Logarithm Properties: Expanding and Condensing★★★☆☆⏱ 4 min

Because logarithms are inverses of exponents, their properties directly correspond to familiar exponent rules. For any positive base $b \neq = 1$ , positive arguments $M, N$ , and real number $k$ , the three core properties are used for two common AP exam tasks: expanding a single condensed expression into simpler terms, or condensing a sum/difference of logs into a single expression.

Product Rule: $lo g_{b} (M N) = lo g_{b} M + lo g_{b} N$ (corresponds to $b^{m} b^{n} = b^{m + n}$ )
Quotient Rule: $lo g_{b} (\frac{M}{N}) = lo g_{b} M - lo g_{b} N$ (corresponds to $\frac{b ^{m}}{b ^{n}} = b^{m - n}$ )
Power Rule: $lo g_{b} (M^{k}) = k lo g_{b} M$ (corresponds to $(b^{m})^{k} = b^{mk}$ )

📐 Worked Example

Expand $lo g_{3} (\frac{9 x ^{5}}{y})$ fully, where $x > 0, y > 0$ .

1
Apply the quotient rule first to split the fraction:
2
$\log_3\left(\frac{9x^5}{\sqrt{y}}\right) = \log_3(9x^5) - \log_3(\sqrt{y})$
3
Apply the product rule to the first term, then rewrite the square root as an exponent:
4
$\log_3 9 + \log_3(x^5) - \log_3(y^{1/2})$
5
Evaluate the constant log term and apply the power rule to variable terms: $lo g_{3} 9 = 2$ , so the fully expanded form is:
6
$2 + 5\log_3 x - \frac{1}{2}\log_3 y$

Exam tip:

When condensing expressions, always move coefficients inside logarithms as exponents before combining terms with product/quotient rules—this avoids common coefficient errors.

3. Change of Base Formula★★★☆☆⏱ 3 min

The change of base formula rewrites a logarithm of any base as a ratio of logarithms with another base. This is required to evaluate any non-common/non-natural logarithm with a calculator, and to simplify expressions with mixed bases.

📘 Definition

Change of Base Formula

$lo g_{b} x = \frac{l o g _{a} x}{l o g _{a} b}$

For any positive $a, b \neq = 1$ , and positive $x$ , this formula holds. On the AP exam, use $a = 10$ (common log) or $a = e$ (natural log) for calculator evaluation.

📐 Worked Example

Simplify $\frac{l o g _{2} 27}{l o g _{2} 9}$ to a single constant.

1
The expression matches the right-hand side of the change of base formula, with $a = 2$ , $b = 9$ , $x = 27$ . This simplifies directly to:
2
$\log_9 27$
3
Rewrite 27 and 9 as powers of the same base 3:
4
$\log_{3^2} 3^3$
5
Apply change of base again to rewrite in base 3:
6
$\frac{\log_3 3^3}{\log_3 3^2} = \frac{3 \log_3 3}{2 \log_3 3}$
7
Cancel $lo g_{3} 3 = 1$ to get the final result:
8
$\frac{3}{2}$

Exam tip:

When asked to evaluate a logarithm where both the base and argument are powers of a common smaller base, always rewrite to cancel logs and avoid unnecessary calculator work.

4. AP-Style Worked Practice★★★★☆⏱ 4 min

✓ Quick check

Test your understanding with this multiple-choice question:

Which of the following is equivalent to $2 lo g 6 - lo g 3$ ?
- $lo g 3$
- $lo g 8$
- $lo g 12$
- $lo g 36$
Reveal answer
2 —
Correct! You applied the power rule first to move the coefficient, then used the quotient rule to combine the terms.

📐 Worked Example

Let $x > 0, y > 0$ , and let $A = lo g_{2} (16 x^{3} y^{- 2})$ . (a) Expand $A$ fully; (b) Calculate $A$ given $lo g_{2} x = 4$ and $lo g_{2} y = - 1$ ; (c) Rewrite $A$ in terms of $ln x$ and $ln y$ .

1
Part (a): Apply product rule to split the argument, then power rule and evaluate the constant log:
2
$\log_2 16 + \log_2 x^3 + \log_2 y^{-2} = 4 + 3\log_2 x - 2\log_2 y$
3
Part (b): Substitute the given values:
4
$A = 4 + 3(4) - 2(-1) = 4 + 12 + 2 = 18$
5
Part (c): Use change of base to convert base 2 logs to natural logs, $lo g_{2} x = \frac{l n x}{l n 2}$ and $lo g_{2} y = \frac{l n y}{l n 2}$ :
6
$A = 4 + \frac{3\ln x - 2\ln y}{\ln 2}$

📐 Worked Example

The pH of a solution is given by $pH = - lo g_{10} [H^{+}]$ , where $[H^{+}]$ is hydrogen ion concentration in mol/L. Orange juice has $[H^{+}] = 3.2 \times 1 0^{- 4}$ mol/L. Calculate its pH to one decimal place. Milk has pH 6.6, find its $[H^{+}]$ in scientific notation.

1
For orange juice, substitute into the formula and expand using product rule:
2
$\text{pH} = -\log_{10}(3.2 \times 10^{-4}) = -\left(\log_{10} 3.2 + \log_{10} 10^{-4}\right) = -(0.505 - 4) \approx 3.5$
3
For milk, rearrange and convert to exponential form:
4
$6.6 = -\log_{10} [\text{H}^+] \implies [\text{H}^+] = 10^{-6.6} = 10^{0.4} \times 10^{-7} \approx 2.5 \times 10^{-7}$

5. Common Pitfalls

Wrong move:

$lo g_{b} (M + N) = lo g_{b} M + lo g_{b} N$

Why:

Students confuse the product rule for logs with addition inside the log; no general rule exists for the logarithm of a sum.

Correct move:

Only apply logarithm rules to product, quotient, and power operations inside the log, never addition or subtraction.

Wrong move:

$\frac{l o g _{b} M}{l o g _{b} N} = lo g_{b} (\frac{M}{N})$

Why:

Students confuse the change of base formula (quotient of two separate logs) with the quotient rule (which applies to a quotient inside a single log).

Correct move:

Only apply the quotient rule when division is inside the logarithm, not between two separate logarithms.

Wrong move:

$lo g_{b} (- x) = - lo g_{b} x$ for $x > 0$

Why:

Students incorrectly apply the power rule to a negative argument, forgetting the domain requirement that all logarithm arguments must be positive.

Correct move:

Always check domain first: $lo g_{b} (- x)$ is only defined for $x < 0$ , and a negative sign in the argument can never be pulled out as a coefficient.

Wrong move:

$(lo g_{b} M)^{k} = k lo g_{b} M$

Why:

Students misapply the power rule, which only applies to a power inside the logarithm, not to the entire logarithm raised to a power.

Correct move:

Only use the power rule when the exponent is on the argument of the log, not on the log itself.

Wrong move:

$lo g_{b} (\frac{1}{x}) = \frac{1}{l o g _{b} x}$

Why:

Students incorrectly apply reciprocal rules from general algebra to logarithms, forgetting the power rule for reciprocals.

Correct move:

Rewrite $\frac{1}{x}$ as $x^{- 1}$ , then apply the power rule to get $lo g_{b} (\frac{1}{x}) = - lo g_{b} x$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Core Definition	$y = lo g_{b} x ⟺ b^{y} = x$	$b > 0, b \neq = 1, x > 0$
Common Logarithm	$lo g x = lo g_{10} x$	No base written explicitly; used for pH, Richter scale
Natural Logarithm	$ln x = lo g_{e} x$	Base $e \approx 2.71828$ ; used for continuous growth
Product Rule	$lo g_{b} (M N) = lo g_{b} M + lo g_{b} N$	$M > 0, N > 0$ ; works for any number of factors
Quotient Rule	$lo g_{b} (\frac{M}{N}) = lo g_{b} M - lo g_{b} N$	$M > 0, N > 0$
Power Rule	$lo g_{b} (M^{k}) = k lo g_{b} M$	$M > 0$ ; works for any real $k$
Reciprocal Rule	$lo g_{b} (\frac{1}{M}) = - lo g_{b} M$	Special case of power rule with $k = - 1$
Change of Base Formula	$lo g_{b} x = \frac{l o g _{a} x}{l o g _{a} b}$	Use $a = 10$ or $a = e$ for calculator evaluation
Logarithm of Base	$lo g_{b} b = 1$	Follows directly from core definition
Logarithm of 1	$lo g_{b} 1 = 0$	Follows directly from core definition

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
Expand condensed logarithmic expression
2023 · AP Precalculus
Evaluate ratio of logarithms

What's Next

Mastery of logarithmic expressions is the foundational prerequisite for all remaining topics in AP Precalculus Unit 2: Exponential and Logarithmic Functions, which makes up 25-30% of your total AP exam score. You will apply these simplification and manipulation techniques to isolate variables when solving equations, model real-world growth and decay, and graph logarithmic functions. Errors in simplifying logarithmic expressions almost always lead to incorrect answers on higher-weight problems, so be sure to master this topic before moving on.