Logarithmic function manipulation

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

1. Core Logarithm Rules for Expansion★★☆☆☆⏱ 4 min

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All logarithm manipulation rules are direct corollaries of exponent rules, since by definition $lo g_{b} (a) = x$ is equivalent to $b^{x} = a$ . The product, quotient, and power rules follow directly from matching exponent properties, and are used to expand complex logarithmic expressions into simpler terms.

🔬 Derivation

Goal:

Derive the three core logarithm algebra rules from exponent properties

Starting from:

Definition of a logarithm and core exponent rules

1
For the product rule: let $x = lo g_{b} (m)$ and $y = lo g_{b} (n)$ . By definition, $b^{x} = m$ and $b^{y} = n$ .
2
Multiply the exponentials: $b^{x} \cdot b^{y} = b^{x + y} = m \cdot n$ . Convert back to logarithmic form to get: $lo g_{b} (mn) = lo g_{b} m + lo g_{b} n$ .
3
For the quotient rule: divide the exponentials to get $\frac{b ^{x}}{b ^{y}} = b^{x - y} = \frac{m}{n}$ . Convert back to get: $lo g_{b} (\frac{m}{n}) = lo g_{b} m - lo g_{b} n$ .
4
For the power rule: raise $b^{x} = m$ to the $k$ power: $(b^{x})^{k} = b^{x k} = m^{k}$ . Convert back to get: $lo g_{b} (m^{k}) = k lo g_{b} m$ .

Result:

All three rules hold for any valid base $b > 0, b \neq = 1$ , and positive arguments $m, n$ .

📐 Worked Example

Expand $lo g_{2} (\frac{4 x ^{3} y - 1}{z ^{2}})$ fully into a sum or difference of constant multiples of simple logarithms, given $x > 0, y > 1, z > 0$ .

1
Apply the quotient rule first to split the fraction:
$\log_2(4x^3 \sqrt{y-1}) - \log_2(z^2)$
2
Apply the product rule to split the first term, and rewrite the square root as a fractional exponent:
$\log_2 4 + \log_2(x^3) + \log_2\left((y-1)^{1/2}\right) - \log_2(z^2)$
3
Simplify the constant term $lo g_{2} 4 = 2$ , then apply the power rule to all terms with exponents:
$2 + 3\log_2 x + \frac{1}{2}\log_2(y-1) - 2\log_2 z$
4
Confirm all arguments of the final expression are positive, so the expansion is valid.

Exam tip:

When expanding, always rewrite roots as fractional exponents before applying the power rule to avoid swapping the exponent value.

2. Condensing Logarithmic Expressions★★☆☆☆⏱ 3 min

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Condensing is the reverse process of expanding: we combine multiple logarithmic terms into a single simplified logarithm. This is most often required before solving logarithmic equations, or when rewriting logarithmic functions to identify key features like intercepts or asymptotes. Always move coefficients to exponents first before combining terms to avoid misapplying rules.

📐 Worked Example

Condense $3 ln x + \frac{1}{2} ln (x + 1) - 2 ln (x - 4)$ into a single logarithm with coefficient 1, given $x > 4$ .

1
Reverse the power rule to move all coefficients to exponents of their arguments:
$\ln x^3 + \ln (x+1)^{1/2} - \ln (x-4)^2$
2
Combine the two positive terms using the product rule:
$\ln\left(x^3 \cdot (x+1)^{1/2}\right) - \ln (x-4)^2$
3
Combine the difference using the quotient rule to get the final single logarithm:
$\ln\left(\frac{x^3 \sqrt{x+1}}{(x-4)^2}\right)$
4
Confirm the argument of the final log is positive for $x > 4$ , matching the domain of the original expression.

Exam tip:

Never add coefficients of logs with different arguments. Always move coefficients to exponents first before combining any terms.

3. Change of Base Formula★★★☆☆⏱ 3 min

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The change of base formula allows us to rewrite a logarithm of any base into a ratio of logarithms with a new base of our choice. On the AP exam, this is used to evaluate non-standard base logarithms with a calculator, or to convert all terms in an expression to the same base for further manipulation.

🔬 Derivation

Goal:

Derive the general change of base formula

Starting from:

Definition of a logarithm

1
Let $y = lo g_{b} a$ , so by definition $b^{y} = a$ .
2
Take $lo g_{k}$ of both sides for any valid new base $k > 0, k \neq = 1$ :
$y \log_k b = \log_k a$
3
Solve for $y$ to get the general formula.

Result:

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

📐 Worked Example

Evaluate $lo g_{6} 12$ to three decimal places, and write the exact value as a ratio of natural logarithms.

1
Apply the change of base formula with $k = e$ (natural log) to get the exact form:
$\log_6 12 = \frac{\ln 12}{\ln 6}$
2
Use a calculator to find the approximate values of the natural logs: $ln 12 \approx 2.4849$ and $ln 6 \approx 1.7918$ .
3
Divide the numerator by the denominator to get the final approximation:
$\frac{2.4849}{1.7918} \approx 1.387$
4
Verify by checking $6^{1.387} \approx 12$ , which confirms the result is correct.

Exam tip:

Always double-check the order of numerator and denominator: original argument goes in the numerator, original base goes in the denominator. Swapping gives the reciprocal of the correct answer.

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

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📐 Worked Example

Which of the following is equivalent to $2 lo g_{5} 10 - lo g_{5} 4$ ? A) 1 B) 2 C) $lo g_{5} 2$ D) $2 lo g_{5} 6$

1
Apply reverse power rule to move the coefficient to the exponent:
$2\log_5 10 = \log_5 10^2 = \log_5 100$
2
Apply the quotient rule to combine the two terms:
$\log_5 100 - \log_5 4 = \log_5\left(\frac{100}{4}\right) = \log_5 25$
3
Simplify using $lo g_{b} b^{k} = k$ , so $lo g_{5} 25 = lo g_{5} 5^{2} = 2$ . The correct answer is B.

📐 Worked Example

Let $f (x) = lo g_{2} (\frac{8 x ^{2}}{x + 3})$ , defined for $x > 0$ . (a) Expand $f (x)$ fully (b) Condense back to verify (c) Rewrite in terms of natural logs

1
(a) Apply rules sequentially to get the full expansion:
$f(x) = 3 + 2\log_2 x - \log_2 (x+3)$
2
(b) Reverse the expansion steps to recover the original function:
$\log_2 8 + \log_2 x^2 - \log_2 (x+3) = \log_2\left(\frac{8x^2}{x+3}\right)$
3
(c) Two equivalent correct forms using change of base:
$f(x) = \frac{\ln\left(\frac{8x^2}{x+3}\right)}{\ln 2} \quad \text{or} \quad f(x) = 3 + \frac{2\ln x}{\ln 2} - \frac{\ln(x+3)}{\ln 2}$

📐 Worked Example

Richter magnitude $R = \frac{l n I - l n I _{0}}{l n 10}$ . If $I = 1 0^{5.9} I_{0}$ , find $R$ .

1
Substitute $I$ into the formula:
$R = \frac{\ln\left(10^{5.9} I_0\right) - \ln I_0}{\ln 10}$
2
Apply product rule and cancel terms:
$R = \frac{\ln 10^{5.9} + \ln I_0 - \ln I_0}{\ln 10}$
3
Apply power rule and simplify:
$R = \frac{5.9 \ln 10}{\ln 10} = 5.9$

5. Common Pitfalls

Wrong move:

Writing $lo g_{b} (m + n) = lo g_{b} m + lo g_{b} n$

Why:

Students confuse the product rule for $lo g_{b} (mn)$ with a sum inside the logarithm. No general rule exists for sums of arguments.

Correct move:

Leave $lo g_{b} (m + n)$ unmodified unless you can factor the argument into a product to apply the product rule.

Wrong move:

Writing $\frac{l o g _{b} a}{l o g _{b} c} = lo g_{b} a - lo g_{b} c$

Why:

Students confuse the change of base ratio of two logs with the quotient rule for a division inside a single log.

Correct move:

Only apply the quotient rule when the entire fraction is the argument of a single logarithm, not when two logarithms are divided.

Wrong move:

Rewriting $ln x^{2}$ as $2 ln x$ without an absolute value

Why:

$ln x^{2}$ is defined for all $x \neq = 0$ , but $2 ln x$ is only defined for $x > 0$ , so domains do not match.

Correct move:

When applying the power rule to an even power, write $2 ln ∣ x ∣$ to preserve the original domain.

Wrong move:

Writing $lo g_{b} (m^{k}) = (lo g_{b} m)^{k}$

Why:

Students misinterpret the power rule, confusing the exponent on the argument with an exponent on the entire logarithm.

Correct move:

The power rule moves the exponent on the argument out front as a scalar multiplier, never as an exponent on the log.

Wrong move:

Condensing $lo g_{2} x + lo g_{3} y$ into $lo g_{2} (x y)$

Why:

Students forget that all core logarithm combination rules require terms to have the same base.

Correct move:

Use change of base to convert both logs to the same base before attempting to combine them.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Product Rule	$lo g_{b} (mn) = lo g_{b} m + lo g_{b} n$	Requires $b > 0, b \neq = 1$ , $m > 0$ , $n > 0$ ; same base only
Quotient Rule	$lo g_{b} (\frac{m}{n}) = lo g_{b} m - lo g_{b} n$	Requires $b > 0, b \neq = 1$ , $m > 0$ , $n > 0$ ; same base only
Power Rule	$lo g_{b} (m^{k}) = k lo g_{b} m$	Requires $b > 0, b \neq = 1$ , $m > 0$ ; use $k lo g_{b} ∣ m ∣$ for even $k$
Change of Base	$lo g_{b} a = \frac{l o g _{k} a}{l o g _{k} b}$	Any valid base $k$ ; use $k = e$ or $k = 10$ for calculation
Reverse Power Rule	$k lo g_{b} a = lo g_{b} (a^{k})$	First step for all condensing problems
Logarithm of 1	$lo g_{b} 1 = 0$	True for all valid bases $b$
Logarithm of Base	$lo g_{b} b = 1$	True for all valid bases $b$

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.

2025 · MCQ
Expand logarithmic expression
2024 · FRQ
Rewrite via change of base

What's Next

Logarithmic manipulation is the foundational skill for almost all remaining topics in Unit 2, and it is used heavily across the rest of AP Precalculus. You will use these manipulation skills to solve exponential and logarithmic equations, analyze the key features of logarithmic functions, and model real-world exponential growth and decay problems, all of which are heavily tested on the AP exam. Incorrect manipulation leads to frequent errors in domain, solution sets, and graph features, so mastering this topic is critical for exam success.