Solving exponential equationsAQA A-Level Maths: Revision notes

An exponential equation has the unknown in the index, such as $a^x=b$ with $a>0$, $a\neq1$ and $b>0$. If both sides can be written as powers of the same base you can compare indices directly: $2^x=32=2^5$ gives $x=5$. Most equations cannot be written that way, for example $5^x=40$, so you use logarithms instead. The equation has a real solution only when $b>0$, because $a^x$ is always positive.

Take logarithms of both sides, using base 10 ($\lg$) or base e ($\ln$). Then use the power law $\log(a^x)=x\log a$ to bring the index down: $a^x=b\;\Rightarrow\;x\lg a=\lg b\;\Rightarrow\;x=\frac{\lg b}{\lg a}.$ The same works with $\ln$: $x=\frac{\ln b}{\ln a}$. Both give the same value. Example: $5^x=40$ gives $x=\frac{\lg40}{\lg5}=\frac{1.6021}{0.69897}=2.29$ (3 s.f.). Check: $5^{2.29}\approx40$.

If the index is $kx+c$, take logarithms and then solve the linear equation. Example: $4^{3x-1}=100$ gives $(3x-1)\lg4=\lg100=2$, so $3x-1=\frac{2}{\lg4}=3.32$ and $x=1.44$. Isolate the power first if it has a coefficient: $3\mathrm{e}^{2x}=21$ becomes $\mathrm{e}^{2x}=7$, then $2x=\ln7$, so $x=\frac12\ln7=0.973$.

When the base is $\mathrm{e}$, use the natural logarithm because $\ln\mathrm{e}^x=x$ exactly. So $\mathrm{e}^x=b$ gives $x=\ln b$ with no division. For any other base you can still use $\ln$: $a^x=b$ gives $x=\frac{\ln b}{\ln a}$. Exact answers are written as logarithms, such as $x=\frac12\ln7$; give a decimal only when the question asks for 3 s.f. or similar, and keep the full calculator value until the last step.

Growth and decay models give equations of the form $a^t=b$. For example, £2000 at 3.5% compound interest is worth $2000\times1.035^n$ after $n$ years. To reach £5000: $1.035^n=2.5$, so $n=\frac{\lg2.5}{\lg1.035}=26.6$. Because the value must exceed £5000 after a whole number of years, the answer is 27 years, not 26.6. Doubling time: solve $a^t=2$, so $t=\frac{\lg2}{\lg a}$, which does not depend on the starting amount. For decay, $0.85^t=0.1$ gives $t=\frac{\lg0.1}{\lg0.85}=14.2$; both logarithms are negative, so the quotient is positive.