Math Core

How long until an investment doubles? When will a cooling cup of coffee reach drinking temperature? Questions like these ask for an unknown exponent, and the tool for pulling an exponent down is the logarithm. This lesson collects the strategies for solving exponential and logarithmic equations, along with the checks that keep you from reporting solutions that don't exist.

The one-to-one property

Both $b^x$ and $\log_b x$ are one-to-one functions: different inputs give different outputs. So for $b > 0$ , $b \ne 1$ :

b^{u} = b^{v} \iff u = v, \qquad\qquad \log_b u = \log_b v \iff u = v \quad (u, v > 0).

If you can write both sides of an equation with the same base, you can simply set the exponents (or the arguments) equal.

Worked example: Rewrite with a common base

Solve $8^{x - 1} = 4^{x + 2}$ .

Solution. Both $8$ and $4$ are powers of $2$ :

\left(2^3\right)^{x - 1} = \left(2^2\right)^{x + 2} \quad\Longrightarrow\quad 2^{3x - 3} = 2^{2x + 4}.

Set the exponents equal: $3x - 3 = 2x + 4$ , so $x = 7$ .

Check: $8^{6} = 262{,}144$ and $4^{9} = 262{,}144$ . ✓

Taking the log of both sides

Most equations don't have a convenient common base. Then the strategy is: isolate the exponential expression, take a logarithm of both sides, and use the power property to bring the exponent down.

Solving exponential equations

Isolate the exponential expression, like $b^{(\ldots)} = c$ .
Take $\ln$ (or $\log$ ) of both sides.
Use $\ln\left(b^{u}\right) = u\ln b$ and solve the resulting linear equation.
Give the exact answer in terms of logs, then a decimal approximation.

If the base is $e$ , use $\ln$ : since $\ln e^{u} = u$ , the exponent comes down with no extra factor.

Worked example: Isolate, then take ln

Solve $3e^{2x} + 1 = 25$ . Give an exact answer and a decimal to three places.

Solution. Isolate the exponential:

3e^{2x} = 24 \quad\Longrightarrow\quad e^{2x} = 8.

Take $\ln$ of both sides: $2x = \ln 8$ , so

x = \frac{\ln 8}{2} \approx 1.040.

Since $\ln 8 = 3\ln 2$ , you could also write $x = \dfrac{3\ln 2}{2}$ .

Common mistake

Isolate the exponential before taking logs. Writing $\ln\left(3e^{2x} + 1\right) = \ln 25$ is true but useless, because there is no property for the log of a sum. And don't "take the log" of just one term: $\ln\left(3e^{2x}\right) + \ln 1$ is not the log of the left side.

When the variable appears in exponents on both sides with different bases, take logs and collect the $x$ -terms.

Worked example: Different bases on each side

Solve $2^{x + 1} = 5^{x}$ .

Solution. Take $\ln$ of both sides and bring down the exponents:

(x + 1)\ln 2 = x\ln 5.

This is a linear equation in $x$ ; $\ln 2$ and $\ln 5$ are just constants. Distribute and collect:

x\ln 2 + \ln 2 = x\ln 5 \quad\Longrightarrow\quad \ln 2 = x(\ln 5 - \ln 2) \quad\Longrightarrow\quad x = \frac{\ln 2}{\ln 5 - \ln 2} = \frac{\ln 2}{\ln 2.5} \approx 0.756.

Check: $2^{1.756} \approx 3.38$ and $5^{0.756} \approx 3.38$ . ✓

Equations of quadratic type

An equation such as $e^{2x} - 5e^{x} + 6 = 0$ contains $e^{2x} = \left(e^{x}\right)^2$ . Substituting $u = e^x$ turns it into a quadratic.

Worked example: Substitution

Solve $e^{2x} - 5e^{x} + 6 = 0$ .

Solution. Let $u = e^{x}$ . Then $u^2 - 5u + 6 = 0$ , which factors as $(u - 2)(u - 3) = 0$ . So $u = 2$ or $u = 3$ :

e^{x} = 2 \;\Longrightarrow\; x = \ln 2 \approx 0.693, \qquad e^{x} = 3 \;\Longrightarrow\; x = \ln 3 \approx 1.099.

Both are valid. If one of the $u$ -values had been zero or negative, it would give no solution, since $e^{x} > 0$ for every $x$ .

Solving logarithmic equations

For equations with logarithms, the idea runs the other way: get a single logarithm on one side, then rewrite in exponential form.

Solving logarithmic equations

Use the properties to condense each side to a single log.
If the equation is $\log_b(\ldots) = c$ , rewrite it as $(\ldots) = b^{c}$ . If it is $\log_b(\ldots) = \log_b(\ldots)$ , set the arguments equal.
Solve the resulting algebraic equation.
Check every solution in the original equation. Reject any value that makes an argument zero or negative.

Step 4 is not optional. Condensing $\log_b u + \log_b v$ into $\log_b(uv)$ enlarges the domain: $uv$ can be positive when $u$ and $v$ are both negative, even though the original logs are undefined there. That is how extraneous solutions appear.

Worked example: An extraneous solution

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

Solution. Condense, then convert to exponential form:

\log_2\left(x(x - 2)\right) = 3 \quad\Longrightarrow\quad x(x - 2) = 2^3 = 8.

Solve the quadratic: $x^2 - 2x - 8 = 0$ , so $(x - 4)(x + 2) = 0$ and $x = 4$ or $x = -2$ .

Check: $x = 4$ gives $\log_2 4 + \log_2 2 = 2 + 1 = 3$ . ✓ But $x = -2$ makes $\log_2(-2)$ undefined, so it is extraneous.

The only solution is $x = 4$ .

Tip

Before solving a log equation, write down its domain (every argument positive). Here that was $x > 2$ . Then you can reject extraneous answers at a glance.