Solving exponential and logarithmic equations

When both sides share a base

To solve $2^x = 32$, notice that $32 = 2^5$. Then $2^x = 2^5$, and equal powers of the same base have equal exponents, so $x = 5$.

The answer need not be a positive whole number. $2^x = \frac{1}{8}$ is $2^x = 2^{-3}$, so $x = -3$ (lesson 1.4.1). And $4^x = 8$ becomes $2^{2x} = 2^3$ once both sides are written in base 2, so $2x = 3$ and $x = \frac{3}{2}$.

That works when you can spot the base. Usually you cannot: try writing 20 as a power of 3. For that you need logarithms.

A solution is a crossing point

Solving $3^x = c$ means finding where the curve $y = 3^x$ meets the horizontal line $y = c$.

A table brackets the crossing. $3^2 = 9$ and $3^3 = 27$, so $x$ is between 2 and 3. Narrowing the bracket, as bisection does in lesson 1.3.2, closes in on it:

So the solution lies between 2.72 and 2.73. The next section finds it exactly.

The picture also shows when there is no solution. $3^x$ is always positive (lesson 1.4.2), so it never equals 0 or a negative number. And it always rises, so it meets each line above the axis exactly once.

The general method: take logs

Here is $3^x = 20$ solved step by step. Before you press Why?, try to say which rule each step uses.

1. 1
   $$3^x = 20$$

   Why?
2. 2
   $$\ln\left(3^x\right) = \ln 20$$

   Why?
3. 3
   $$x \ln 3 = \ln 20$$

   Why?
4. 4
   $$x = \frac{\ln 20}{\ln 3} \approx 2.7268$$

   Why?

That matches the table: between 2.72 and 2.73.

The signs take care of themselves. For $0.5^x = 0.1$, both logs are negative: $x = \frac{\ln 0.1}{\ln 0.5} \approx 3.3219$, a positive answer. For $2^x = 0.3$, the target is below 1, so $x = \frac{\ln 0.3}{\ln 2} \approx -1.7370$ is negative.

Isolate the power first

Solve $5 \cdot 2^x + 3 = 43$. Taking logs straight away gets you stuck: $\ln(5 \cdot 2^x + 3)$ has no rule, because of the $+\,3$. Undo the arithmetic around the power first.

With base $e$, $\ln$ undoes the power directly. Solve $4e^{2x} = 100$:

Isolating can also reveal that there is no solution. $2 \cdot 5^x + 9 = 3$ becomes $5^x = -3$, and no power of 5 is negative.

The unknown in two exponents

Solve $2^{x+1} = 5^x$. Taking logs works here too, and then it becomes an ordinary linear equation.

1. 1
   $$\ln\left(2^{x+1}\right) = \ln\left(5^x\right)$$

   Why?
2. 2
   $$(x + 1)\ln 2 = x \ln 5$$

   Why?
3. 3
   $$\begin{aligned} \ln 2 &= x \ln 5 - x \ln 2 \\ &= x(\ln 5 - \ln 2) \end{aligned}$$

   Why?
4. 4
   $$x = \frac{\ln 2}{\ln 5 - \ln 2} \approx 0.7565$$

   Why?

Equations that hide a quadratic

Solve $e^{2x} - 5e^x + 6 = 0$. Taking logs would get stuck on the sums. Instead, notice that $e^{2x} = (e^x)^2$ by the power-of-a-power rule, and give $e^x$ a name.

Now undo the substitution. $e^x = 2$ gives $x = \ln 2 \approx 0.6931$, and $e^x = 3$ gives $x = \ln 3 \approx 1.0986$. Two solutions.

Change one sign and a root disappears. $e^{2x} - e^x - 6 = 0$ factors as $(u - 3)(u + 2) = 0$, so $u = 3$ or $u = -2$. But $u = e^x$ is never negative, so only $x = \ln 3$ survives.

Solving log equations

To free an unknown from inside a log, apply the matching power to both sides. This undoes the log.

A negative log is no obstacle: $\ln x = -2$ gives $x = e^{-2} \approx 0.1353$, a positive number below 1 (lesson 1.4.3).

When there are two logs, combine them first. For $\log_2 x + \log_2(x - 2) = 3$, the product rule gives one log, and undoing it leaves a quadratic:

So $x = 4$ or $x = -2$. Check both in the original equation. $x = 4$ gives $\log_2 4 + \log_2 2 = 2 + 1 = 3$. But $x = -2$ would need $\log_2(-2)$, which does not exist, so the only solution is $x = 4$.

Sometimes nothing survives. Solve $\ln(x - 5) = \ln(2x)$. Equal logs mean equal inputs, so $x - 5 = 2x$, which gives $x = -5$. But then $x - 5 = -10$, and $\ln(-10)$ does not exist. The equation has no solution.

Let SymPy check your algebra

The last line is a warning. SymPy returns $-5$ for $\ln(x - 5) = \ln(2x)$, because it allows logs of negative numbers as complex numbers. So check every answer yourself, even one from SymPy.

Common mistakes

1. Taking logs before isolating the power. $\ln(5 \cdot 2^x + 3)$ has no rule. Undo the $+\,3$ and the $5$ first.
2. Simplifying the answer wrongly. $\frac{\ln 20}{\ln 3}$ is not $\ln\frac{20}{3}$, and not $\ln 20 - \ln 3$.
3. Expecting a solution for every target. $b^x = c$ has none when $c \le 0$.
4. Rejecting a negative log. $\ln x = -2$ is fine: $x = e^{-2}$.
5. Keeping every root of the quadratic. $e^x = -2$ and $\log_2(-2)$ both fail; check each root.
6. Dropping brackets. $\ln 2^{x+1}$ is $(x + 1)\ln 2$, not $x + 1 \cdot \ln 2$.