6.4 Solving equations of the form a^x = b — Practise with help

Work each question closed book first. Isolate the exponential, then take logarithms of both sides.

Warm-up: remember

Work out $\ln e^2$ .
Write $3\ln x$ as a single logarithm.
Solve $2^x = 8$ by writing both sides as powers of $2$ .

Answers

$\ln e^2 = 2$ , because the logarithm undoes the exponential.
$3\ln x = \ln x^3$ .
$2^x = 2^3$ , so $x = 3$ .

If question 2 was hard, revise the power law in Section 6.3 before you continue.

Practise with help

1. Solve $3^x = 40$ , giving $x$ exactly and to three significant figures.

Hint

Take the natural logarithm of both sides. The power law brings the $x$ to the front.

Next step

Write $x\ln 3 = \ln 40$ , then divide both sides by $\ln 3$ .

Full working

x = \frac{\ln 40}{\ln 3} = 3.357\ldots \approx 3.36.

The exact answer keeps both logarithms; the decimal comes last.

2. Solve $e^x = 12$ , giving $x$ exactly.

Hint

The natural logarithm undoes the exponential. No change of base is needed.

Next step

Take $\ln$ of both sides. The left side becomes $x$ .

Full working

x = \ln 12 \approx 2.485.

Because $\ln 12 > 0$ , the value of $e^x$ is greater than $1$ , which matches the positive exponent.

3. Solve $e^{3x} = 20$ , giving $x$ exactly.

Hint

Take logarithms first, then divide by the coefficient $3$ .

Next step

Taking $\ln$ gives $3x = \ln 20$ . Divide both sides by $3$ .

Full working

x = \frac{\ln 20}{3} \approx 0.999.

Dividing by $3$ is the last step, after the logarithm has isolated the exponent.

4. Solve $2^{2x} - 9\left(2^x\right) + 8 = 0$ .

Hint

Write $2^{2x}$ as $\left(2^x\right)^2$ and let $u = 2^x$ . The equation becomes a quadratic in $u$ .

Next step

Factorise $u^2 - 9u + 8 = 0$ and solve for $u$ . Then substitute back to find $x$ .

Full working

u^2 - 9u + 8 = (u - 1)(u - 8) = 0,

so $u = 1$ or $u = 8$ . Therefore $2^x = 1$ , giving $x = 0$ , or $2^x = 8$ , giving $x = 3$ . Both values are valid.

5. Solve $4^{x+1} = 50$ , giving $x$ to three significant figures.

Hint

Take logarithms: $(x + 1)\ln 4 = \ln 50$ . The exponent $x + 1$ stays together until the end.

Next step

Make $x + 1$ the subject, then subtract $1$ .

Full working

x + 1 = \frac{\ln 50}{\ln 4}, \qquad x = \frac{\ln 50}{\ln 4} - 1 = 1.822\ldots \approx 1.82.

The $-1$ comes from the exponent $x + 1$ .

6. Solve $\log_2\left(x - 1\right) = 3$ , and state the domain.

Hint

A logarithm needs a positive input, so $x - 1 > 0$ . Rewrite the equation as a power with base $2$ .

Next step

Write $x - 1 = 2^3 = 8$ , then solve for $x$ and check the domain.

Full working

The domain is $x > 1$ . Rewriting the logarithm gives

x - 1 = 8, \qquad x = 9.

The answer $9$ is greater than $1$ , so it lies in the domain.

7. Explain why $5^x = -2$ has no real solution.

Hint

Think about the graph of $y = 5^x$ , and the values it can take.

Next step

Decide whether $5^x$ can ever be negative, and what the graph does for very large negative $x$ .

Full working

The graph of $y = 5^x$ lies above the $x$ -axis for every real $x$ . As $x$ becomes very negative the curve approaches $0$ , but it never reaches a negative value. So the value $-2$ is outside the range, and the equation has no real solution.