6.4 Solving equations of the form a^x = b — Learn

Prerequisites: the laws of logarithms from Section 6.3; solving a quadratic by factorising from Chapter 2; the domain of a logarithm.

By the end of this section you can: solve $a^x = b$ by taking logarithms; solve an equation where the exponent is a linear expression; solve a quadratic equation in $e^x$ or $a^x$; solve a simple logarithmic equation; and check each answer against the domain.

1. Take logarithms of both sides

To solve $a^x = b$, take the logarithm of both sides. The power law then brings the $x$ down from the exponent, where it can be handled.

Starting from $a^x = b$ and taking natural logarithms:

This is the change-of-base rule in action: $x = \log_a b$. Any base of logarithm works, but $\ln$ is the usual choice.

Example 1. Solve $3^x = 20$, giving $x$ exactly and to three significant figures.

Take logarithms: $x\ln 3 = \ln 20$, so

Example 2. Solve $e^{2x} = 7$.

The natural logarithm undoes the exponential directly:

Divide by the coefficient of $x$ only at the end.

Example 3. Solve $5^{2x-1} = 30$.

Take logarithms of both sides:

Make $2x - 1$ the subject, then solve for $x$:

2. An equation that is a quadratic in disguise

When an equation contains both $a^{2x}$ and $a^{x}$, write $a^{2x} = \left(a^{x}\right)^2$ and treat $a^x$ as a single unknown. Factorise, then solve each exponential equation.

Example 4. Solve $e^{2x} - 5e^x + 6 = 0$.

Let $u = e^x$. The equation becomes $u^2 - 5u + 6 = 0$, which factorises:

So $e^x = 2$ or $e^x = 3$, giving $x = \ln 2 \approx 0.693$ or $x = \ln 3 \approx 1.099$. Both values are valid, because $e^x$ is positive for every $x$.

Example 5. Solve $2^{2x} - 5\left(2^x\right) + 4 = 0$.

Write $2^{2x} = \left(2^x\right)^2$ and let $u = 2^x$:

So $2^x = 1$, giving $x = 0$, or $2^x = 4$, giving $x = 2$.

3. Solving a logarithmic equation

Reverse the logarithm with its matching exponential. If $\ln x = c$, then $x = e^c$. If $\log_a x = c$, then $x = a^c$. Always check that the answer lies in the domain, because a logarithm needs a positive input.

Example 6. Solve $\ln(x - 2) = 1$.

The input must be positive, so $x > 2$. Exponentiate both sides:

The answer is greater than $2$, so it lies in the domain.

Progress check

Return to the cooling coffee in the chapter problem. Its excess temperature is $E = 60e^{-kt}$, and after $10$ minutes the excess temperature is $45^\circ$C.

1. Substitute $t = 10$ and $E = 45$ to write an equation for $k$.
2. Solve it exactly, and give $k$ to three significant figures.
3. Find the exact time for the excess temperature to fall to $15^\circ$C, and give the time in minutes to one decimal place.
4. Explain why the second time is longer than the first, even though the second temperature fall is smaller.

Watch out. These mistakes are common.

Taking the logarithm of a negative number. The equation $2^x = -8$ has no solution, because $2^x$ is positive for every real $x$. A negative number has no real logarithm.

Dividing by the wrong thing. In $e^{2x} = 7$ the logarithm gives $2x = \ln 7$, so $x = \dfrac{\ln 7}{2}$. Dividing by $7$ or subtracting $2$ gives the wrong answer. Isolate the exponential first.

Stopping at the quadratic. After $u^2 - 5u + 6 = 0$ gives $u = 2$ or $u = 3$, the equation is not solved. Substitute back: $e^x = 2$ and $e^x = 3$ still have to be solved for $x$.

Forgetting the domain. From $\ln(x - 2) = 1$ you get $x = e + 2$. If a candidate made the input zero or negative, it would have to be rejected. Re-check: substitute the answer into the original logarithm and confirm a positive input.

You should now be able to

• solve $a^x = b$ by taking logarithms of both sides;
• solve an equation where the exponent is a linear expression in $x$;
• recognise $a^{2x}$ as $\left(a^x\right)^2$ and solve the resulting quadratic;
• solve a simple equation involving $\ln$ or $\log_a$;
• check each solution against the domain of the logarithm or exponential.