AP Precalculus: Logarithms
Master logarithm properties, conversions, and solving techniques
📚 Understanding Logarithms
Logarithms are the inverse of exponential functions. If you know that \(b^x = y\), then the logarithm tells you what power \(x\) you need. Mastering logarithm properties is essential for solving exponential equations and many AP Precalculus applications.
1 Converting Between Exponential & Logarithmic Form
The logarithm base \(a\) of \(b\) answers the question: "To what power must I raise \(a\) to get \(b\)?" This is the inverse relationship between exponentials and logarithms.
How to Read Logarithms
- \(\log_a b = x\) reads as "log base \(a\) of \(b\) equals \(x\)"
- The base is the small subscript number (\(a\))
- The argument is what's inside the log (\(b\))
- The result is the exponent you need (\(x\))
Exponential → Logarithmic:
\(2^5 = 32 \Rightarrow \log_2 32 = 5\)
\(10^3 = 1000 \Rightarrow \log_{10} 1000 = 3\)
\(e^2 ≈ 7.389 \Rightarrow \ln 7.389 ≈ 2\)
Logarithmic → Exponential:
\(\log_3 81 = 4 \Rightarrow 3^4 = 81\)
\(\log_5 125 = 3 \Rightarrow 5^3 = 125\)
2 Common & Natural Logarithms
Two logarithm bases are used so frequently they have special notation and dedicated calculator buttons.
Calculator button: LOG
Calculator button: LN
\(\log 100 = \log_{10} 100 = 2\) because \(10^2 = 100\)
\(\log 1000 = 3\) because \(10^3 = 1000\)
\(\ln e = 1\) because \(e^1 = e\)
\(\ln e^5 = 5\) because \(e^5 = e^5\)
If your calculator only has LOG and LN buttons, you can compute any base using the change of base formula (see below).
3 Fundamental Logarithm Identities
These identities follow directly from the definition of logarithms and the properties of exponents. They are essential for simplifying and evaluating logarithmic expressions.
\(\log_5 1 = 0\) (any log of 1 is 0)
\(\log_7 7 = 1\) (log of base equals 1)
\(\log_2 2^8 = 8\) (exponent comes out)
\(10^{\log 100} = 100\) (exponent and log cancel)
\(e^{\ln 5} = 5\) (natural log and \(e\) cancel)
For \(\log_a x\): the base \(a > 0\), \(a \neq 1\), and the argument \(x > 0\). You cannot take the log of zero or negative numbers!
4 Properties of Logarithms
These three properties allow you to break apart or combine logarithmic expressions. They mirror the exponent rules: products become sums, quotients become differences, and powers become multipliers.
From: \(a^m \cdot a^n = a^{m+n}\)
From: \(\frac{a^m}{a^n} = a^{m-n}\)
From: \((a^m)^k = a^{mk}\)
Expand: \(\log_2(8x^3y)\)
\(= \log_2 8 + \log_2 x^3 + \log_2 y\) (Product Rule)
\(= 3 + 3\log_2 x + \log_2 y\) (Power Rule + evaluate)
Condense: \(2\log x + \log y - 3\log z\)
\(= \log x^2 + \log y - \log z^3\) (Power Rule)
\(= \log\left(\frac{x^2 y}{z^3}\right)\) (Product & Quotient Rules)
5 Change of Base Formula
The change of base formula converts a logarithm from one base to another. This is essential for calculator use, since most calculators only have LOG (base 10) and LN (base \(e\)) buttons.
Most Common Forms
Using Common Log
\(\log_a b = \frac{\log b}{\log a}\)
Using Natural Log
\(\log_a b = \frac{\ln b}{\ln a}\)
Evaluate: \(\log_3 20\)
\(\log_3 20 = \frac{\log 20}{\log 3} = \frac{1.301}{0.477} ≈ 2.727\)
Or using ln:
\(\log_3 20 = \frac{\ln 20}{\ln 3} = \frac{2.996}{1.099} ≈ 2.727\)
To convert between common and natural log: \(\ln x = \frac{\log x}{\log e} ≈ 2.303 \cdot \log x\)
6 Evaluating Logarithms
To evaluate a logarithm, ask yourself: "What power of the base gives me this argument?" Use the exponential definition \(\log_a b = x \Leftrightarrow a^x = b\).
Strategy for Evaluation
- Identify the base \(a\) and argument \(b\)
- Ask: "What power of \(a\) equals \(b\)?"
- If \(b\) is a power of \(a\), write it as \(a^x\) and the answer is \(x\)
- If not a perfect power, use change of base with a calculator
\(\log_2 32 = ?\) → "2 to what power is 32?" → \(2^5 = 32\) → 5
\(\log_3 \frac{1}{27} = ?\) → \(3^{-3} = \frac{1}{27}\) → −3
\(\log_5 \sqrt{5} = ?\) → \(5^{1/2} = \sqrt{5}\) → 1/2
\(\log_{10} 0.001 = ?\) → \(10^{-3} = 0.001\) → −3
\(\log_2 2 = 1\), \(\log_2 4 = 2\), \(\log_2 8 = 3\), \(\log_2 16 = 4\), \(\log_2 32 = 5\)
\(\log 10 = 1\), \(\log 100 = 2\), \(\log 1000 = 3\)
\(\ln 1 = 0\), \(\ln e = 1\), \(\ln e^2 = 2\)
7 Solving Logarithmic Equations
Logarithmic equations can be solved by converting to exponential form, using logarithm properties to condense, or by applying the one-to-one property.
Methods for Solving
Solve: \(\log_3(x + 2) = 4\)
Convert: \(x + 2 = 3^4 = 81\)
Solve: \(x = 79\)
Check: \(\log_3(79 + 2) = \log_3 81 = 4\) ✓
Solve: \(\log_2(3x - 1) = \log_2(x + 7)\)
Apply one-to-one: \(3x - 1 = x + 7\)
Solve: \(2x = 8 \Rightarrow x = 4\)
Check: \(\log_2(11) = \log_2(11)\) ✓
Solutions that make the argument of any logarithm negative or zero are invalid. Always substitute back to verify.
8 Solving Exponential Equations Using Logarithms
When the variable is in the exponent and you can't rewrite both sides with the same base, take the logarithm of both sides to bring the exponent down.
- Isolate the exponential expression on one side
- Take the log (or ln) of both sides
- Use the power rule: \(\log(a^x) = x \cdot \log a\)
- Solve for the variable
Solve: \(5^{2x-1} = 17\)
Take ln of both sides: \(\ln(5^{2x-1}) = \ln 17\)
Power rule: \((2x-1) \ln 5 = \ln 17\)
Distribute: \(2x \ln 5 - \ln 5 = \ln 17\)
Solve: \(2x \ln 5 = \ln 17 + \ln 5\)
\(x = \frac{\ln 17 + \ln 5}{2 \ln 5} = \frac{\ln 85}{2 \ln 5} ≈ 1.381\)
📋 Quick Reference: Key Formulas
Definition
\(a^x = b \Leftrightarrow \log_a b = x\)
Product Rule
\(\log_a(xy) = \log_a x + \log_a y\)
Quotient Rule
\(\log_a(\frac{x}{y}) = \log_a x - \log_a y\)
Power Rule
\(\log_a(x^k) = k \cdot \log_a x\)
Change of Base
\(\log_a b = \frac{\ln b}{\ln a} = \frac{\log b}{\log a}\)
Inverse Properties
\(\log_a a^x = x\) and \(a^{\log_a x} = x\)