Logarithms: Solving for an Exponent

A logarithm is an exponent

$$2^3 = 8$$

That equation can be read two ways. Given the exponent $3$, it says the result is $8$. Given the result $8$, it says the exponent is $3$. The second reading is a logarithm:

$$\log_2 8 = 3$$

Read it as “log base $2$ of $8$ is $3$” — the power $2$ must be raised to, to give $8$.

$$\log_b x = y \quad\iff\quad b^y = x$$

Exponential form	Logarithmic form
$10^3 = 1000$	$\log_{10} 1000 = 3$
$5^2 = 25$	$\log_5 25 = 2$
$2^{-2} = \tfrac{1}{4}$	$\log_2 \tfrac{1}{4} = -2$
$b^0 = 1$	$\log_b 1 = 0$

Why the logarithm is the inverse of the exponential

The exponential function $f(x) = 2^x$ takes an exponent and returns a result. $\log_2$ takes the result and returns the exponent. Doing one and then the other gets you back where you started:

$$\log_2\left(2^x\right) = x \qquad 2^{\log_2 x} = x$$

That is the definition of inverse functions. So the graph of $y = \log_2 x$ is the graph of $y = 2^x$ reflected in the line $y = x$, and the two swap domain and range:

	$y = 2^x$	$y = \log_2 x$
domain	all real numbers	$x > 0$
range	$y > 0$	all real numbers

This is why the logarithm of zero or a negative number does not exist among the reals. A positive base raised to any power is positive, so no exponent produces $0$ or $-5$.

Solving for an exponent

This is what logarithms are for. When the unknown sits in an exponent, no amount of adding, subtracting, multiplying or dividing can bring it down. A logarithm can.

Solve $3 \cdot 2^t = 48$.

Isolate the power first:

$$2^t = 16$$

Then rewrite in logarithmic form:

$$t = \log_2 16 = 4$$

When the answer is not a whole number, a calculator finishes the job.

Solve $2^x = 20$.

$$x = \log_2 20$$

Evaluating logarithms with a calculator

Calculators have two logarithm keys:

Key	Base	Name
log	$10$	common logarithm
ln	$e \approx 2.718$	natural logarithm

For any other base, the change-of-base formula converts it:

$$\log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}$$ $$\log_2 20 = \frac{\log 20}{\log 2} \approx \frac{1.3010}{0.3010} \approx 4.32$$

Check: $2^{4.32} \approx 20$ ✓ — between $2^4 = 16$ and $2^5 = 32$, as it should be.

Real models: how long until…

Exponential models answer “how much after $t$ years?” Logarithms answer the reverse: “how many years until…?”

$500 grows at $5\%$ a year. How long until it doubles to $1,000?

$$500 \cdot 1.05^t = 1000 \quad\Rightarrow\quad 1.05^t = 2$$ $$t = \log_{1.05} 2 = \frac{\log 2}{\log 1.05} \approx 14.2 \text{ years}$$

A population of $200$ grows as $P = 200e^{0.03t}$. When does it reach $600$?

$$e^{0.03t} = 3 \quad\Rightarrow\quad 0.03t = \ln 3 \quad\Rightarrow\quad t = \frac{\ln 3}{0.03} \approx 36.6$$

The method is always the same: isolate the power, then take the logarithm whose base matches. With base $e$, that is ln; with base $10$, log; with anything else, change of base.

Worked examples

Common mistakes

Practice problems