Exponential and logarithmic equationsEdexcel A-Level Maths: Revision notes

An exponential equation has the unknown in the index. To solve $a^{x}=b$ with $b>0$, take logarithms of both sides (any base, but $\log_{10}$ or $\ln$ is best on a calculator) and use the power rule $\log(a^{x})=x\log a$: $a^{x}=b\;\Rightarrow\;x\log a=\log b\;\Rightarrow\;x=\frac{\log b}{\log a}.$ Example: $5^{x}=40$ gives $x=\frac{\ln40}{\ln5}=2.29$ (3 s.f.). The exact answer is $x=\log_{5}40$, and a negative $x$ is fine when $b<1$, e.g. $4^{x}=0.3$ gives $x=-0.868$.

When the index is an expression such as $3x-1$, the power rule multiplies the whole index by the log. Then rearrange as an ordinary linear equation. Example: $2^{3x-1}=3$. Taking logs: $(3x-1)\log2=\log3$, so $3x-1=\frac{\log3}{\log2}=1.585$, then $3x=2.585$ and $x=0.862$. Always keep full calculator values until the final line, and give the answer to the accuracy asked for (usually 3 s.f.).

If both sides are powers of different bases, take logs, expand the brackets and collect the $x$ terms on one side, then factorise. Example: $3^{x}=2^{x+1}$. Then $x\ln3=(x+1)\ln2$, so $x\ln3-x\ln2=\ln2$, giving $x(\ln3-\ln2)=\ln2$ and $x=\frac{\ln2}{\ln3-\ln2}=1.71$. Alternatively, divide first: $3^{x}=2\times2^{x}$ gives $1.5^{x}=2$ and $x=\frac{\ln2}{\ln1.5}$, the same value.

The change of base formula lets you evaluate a logarithm in any base using your calculator's $\log_{10}$ or $\ln$ keys: $\log_{a}b=\frac{\log_{c}b}{\log_{c}a}.$ So $\log_{5}40=\frac{\ln40}{\ln5}$, which is exactly the answer to $5^{x}=40$. Solving $a^{x}=b$ and writing $x=\log_{a}b$ are the same step, and the formula turns it into a number.

Models such as $V=2500\times1.04^{t}$ lead to $a^{x}=b$ when you are given a value and asked for the time. Set the model equal to the target, isolate the power (divide by the starting value), then take logs. Example: $2500\times1.04^{t}=4000\Rightarrow1.04^{t}=1.6\Rightarrow t=\frac{\ln1.6}{\ln1.04}=12.0$. When asked for a whole number of years, round up if the value must first exceed a target.