Home / Exponential Functions / The inverse function ln and solving $a^x = b$
3.13c · Exponential Functions · Sub-skill
Use ln to undo $e^x$ and to solve equations where the unknown is in the power.
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$\ln x$ (the natural logarithm) is the inverse function of $e^x$: $\ln(e^x) = x$ and $e^{\ln x} = x$. The graph of $y = \ln x$ is the reflection of $y = e^x$ in the line $y = x$. It is only defined for $x > 0$ and passes through $(1, 0)$.
To solve an equation like $e^{0.05t} = 3$, take $\ln$ of both sides: $0.05t = \ln 3$, so $t = \dfrac{\ln 3}{0.05} \approx 21.97$.
For any base $a$, take logarithms of both sides: $x \ln a = \ln b$, so $x = \dfrac{\ln b}{\ln a}$. For example $2^x = 50$ gives $x = \dfrac{\ln 50}{\ln 2} \approx 5.64$.
To find how long it takes to double at a growth multiplier $b$, solve $b^t = 2$: $t = \dfrac{\ln 2}{\ln b}$. The same method with $b^t = 0.5$ gives a half-life.
Solve $2^x = 50$, giving your answer to 2 decimal places.
Take logs: $x \ln 2 = \ln 50$, so $x = \dfrac{\ln 50}{\ln 2} = \dfrac{3.9120}{0.6931} = $ 5.64.
Test yourself
£1000 is invested at 4% compound interest per year.
(a) Write down a formula for the value $V$ after $t$ years.
(b) Find how long it takes for the investment to reach £2000. Give your answer to 1 decimal place.
(c) Find the value of the investment after 18 years. [5 marks]
(a) $V = 1000 \times 1.04^t$. (1 mark)
(b) $1.04^t = 2 \Rightarrow t = \dfrac{\ln 2}{\ln 1.04} = \dfrac{0.6931}{0.03922} = $ 17.7 years. (2 marks)
(c) $V = 1000 \times 1.04^{18} = $ £2025.82. (2 marks)
Practice
Five short questions. Work through them, then reveal the mark scheme to check.