Logarithms: rules, change of base, solving

Q: What is the difference between log and ln?

Both are logarithms with different bases. \log with no base written usually means base 10, the common log. \ln is base e, about 2.718, the natural log. They follow the same rules, and either one works for change of base.

Q: Is log(a + b) equal to log a + log b?

No. The product rule says \log(ab) = \log a + \log b. There is no rule that splits a log of a sum, so leave \log(a + b) as it is. For example, \log(10 + 10) = \log 20 \approx 1.30, but \log 10 + \log 10 = 2.

The key idea

Logs and exponents are inverses. Every log statement is an exponent statement read backwards:

\log_b(x) = y \iff b^y = x \qquad (b > 0,\ b \ne 1,\ x > 0)

The same fact in both forms
Exponential form	Log form
$2^5 = 32$	$\log_2(32) = 5$
$10^{-2} = 0.01$	$\log(0.01) = -2$
$e^0 = 1$	$\ln(1) = 0$

$\log$ with no base means base 10, and $\ln$ means base $e \approx 2.718$ . The three rules come straight from the exponent rules:

\log_b(MN) = \log_b M + \log_b N, \quad \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N, \quad \log_b(M^p) = p\log_b M

Change of base lets any calculator find any log:

\log_b(x) = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}

Worked examples

Example 1: evaluate a log by hand

Problem Find $\log_4(8)$ without a calculator.

Call the answer y. By definition, 4 to the y is 8.
$4^y = 8$
Write both sides as powers of 2.
$(2^2)^y = 2^3$
Same base, so the exponents must match.
$2y = 3$
Divide by 2.
$y = \frac{3}{2}$

Answer $\log_4(8) = \frac{3}{2}$

Example 2: condense into one log

Problem Write $2\log_3(x) + \log_3(5) - \log_3(x + 1)$ as a single logarithm.

Power rule first: the 2 becomes an exponent.
$2\log_3(x) + \log_3(5) - \log_3(x + 1) = \log_3(x^2) + \log_3(5) - \log_3(x + 1)$
Product rule: a sum of logs is the log of a product.
$\log_3(x^2) + \log_3(5) - \log_3(x + 1) = \log_3(5x^2) - \log_3(x + 1)$
Quotient rule: a difference of logs is the log of a quotient.
$\log_3(5x^2) - \log_3(x + 1) = \log_3\left(\frac{5x^2}{x + 1}\right)$

Answer $\log_3\left(\frac{5x^2}{x + 1}\right)$

Example 3: solve an exponential equation

Problem Solve $5^{2x - 1} = 40$ . Round to three decimal places.

40 is not a power of 5, so take the log of both sides and use the power rule to bring the exponent down.
$(2x - 1)\log(5) = \log(40)$
Divide both sides by $\log 5$ .
$2x - 1 = \frac{\log(40)}{\log(5)}$
Add 1 and divide by 2.
$x = \frac{1}{2}\left(1 + \frac{\log(40)}{\log(5)}\right)$
On a calculator, $\frac{\log 40}{\log 5} \approx 2.292$ , so $x \approx \frac{3.292}{2}$ .

Answer $x \approx 1.646$

Example 4 (test-hard): a log equation with an extraneous answer

Problem Solve $\log_2(x) + \log_2(x - 2) = 3$ .

Product rule: combine into one log.
$\log_2(x(x - 2)) = 3$
Rewrite in exponential form: $2^3 = 8$ .
$x(x - 2) = 8$
Expand and set to zero.
$x^2 - 2x - 8 = 0$
Factor.
$(x - 4)(x + 2) = 0$
So $x = 4$ or $x = -2$ . But $\log_2(-2)$ is not defined, so $-2$ is extraneous. Check 4: $\log_2 4 + \log_2 2 = 2 + 1 = 3$ .

Answer $x = 4$

Common mistakes

Splitting a log of a sum. $\log(a + b)$ is not $\log a + \log b$ . Fix: the product rule is for $\log(ab)$ only.
Canceling logs in a fraction. $\frac{\log 40}{\log 5}$ is not $\log 8$ . Fix: divide the two log values; only $\log 40 - \log 5$ equals $\log 8$ .
Moving a coefficient the wrong way. $2\log_3 x = \log_3(x^2)$ , not $\log_3(2x)$ . Fix: the number in front becomes an exponent on the input.
Keeping answers that make a log input negative or zero. Fix: after solving, put each answer into every log in the original equation.
Mixing up the base and the answer. $\log_2 32 = 5$ means $2^5 = 32$ , not $5^2$ or $32^2$ . Fix: say "2 to what power is 32?"

Quick methods

Practice

5 practice questions

What is $\log_2\left(\frac{1}{64}\right)$ ?
1. $-6$
2. $6$
3. $-32$
4. $\frac{1}{6}$
Show answer
Answer: $-6$
$2^6 = 64$ , so $2^{-6} = \frac{1}{64}$ and the log is $-6$ . A fraction less than 1 always gives a negative log when the base is more than 1.
Which expression is equal to $\log(x) + \log(5) - \log(2)$ for $x > 0$ ?
1. $\log\left(\frac{5x}{2}\right)$
2. $\log(5x - 2)$
3. $\log(x + 3)$
4. $\log(10x)$
Show answer
Answer: $\log\left(\frac{5x}{2}\right)$
Adding logs multiplies the inputs and subtracting divides: $\log\left(\frac{x \cdot 5}{2}\right)$ . $\log(5x - 2)$ treats subtracting logs as subtracting inputs, and $\log(10x)$ multiplies by 2 instead of dividing.
Solve $3^x = 20$ . Round to the nearest hundredth.
1. 0.37
2. 1.30
3. 2.73
4. 6.67
Show answer
Answer: 2.73
$x = \frac{\log 20}{\log 3} \approx \frac{1.301}{0.477} \approx 2.73$ . 0.37 divides the logs upside down, 1.30 is just $\log 20$ , and 6.67 is $20 \div 3$ . Check: $3^{2.73} \approx 20$ .
Solve $\log_3(x + 1) = 4$ .
1. $11$
2. $63$
3. $80$
4. $82$
Show answer
Answer: $80$
Exponential form: $x + 1 = 3^4 = 81$ , so $x = 80$ . 63 uses $4^3$ instead of $3^4$ , and 11 uses $3 \cdot 4$ .
Solve $\log_5(x) + \log_5(x - 4) = 1$ .
Show answer
Answer: 5
Combine: $\log_5(x(x - 4)) = 1$ , so $x^2 - 4x = 5$ , which factors as $(x - 5)(x + 1) = 0$ . $x = -1$ makes $\log_5(x)$ undefined, so only $x = 5$ works.

Frequently asked questions

What is the difference between log and ln?

Both are logarithms with different bases. $\log$ with no base written usually means base 10, the common log. $\ln$ is base $e$ , about 2.718, the natural log. They follow the same rules, and either one works for change of base.

Why can't you take the log of a negative number?

A log asks what power of a positive base gives the input. A positive base raised to any real power is always positive, so no real exponent gives 0 or a negative number. That is why answers that make a log input zero or negative get thrown out.

How do I use the change of base formula?

Divide the log of the input by the log of the base, using the same kind of log on top and bottom: $\log_2(10) = \frac{\log 10}{\log 2} \approx 3.32$ . Check by raising: $2^{3.32}$ is about 10.

Is log(a + b) equal to log a + log b?

No. The product rule says $\log(ab) = \log a + \log b$ . There is no rule that splits a log of a sum, so leave $\log(a + b)$ as it is. For example, $\log(10 + 10) = \log 20 \approx 1.30$ , but $\log 10 + \log 10 = 2$ .

The key idea

Worked examples

Common mistakes

Quick methods

Practice

Frequently asked questions

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