Solving equations of the form a^x = bEdexcel International A Level Maths: Revision notes

When the unknown is in the index and the two sides cannot easily be written as powers of one base, take logarithms of both sides and use the power law: $a^x=b\;\Rightarrow\;x\log a=\log b\;\Rightarrow\;x=\frac{\log b}{\log a}.$ Any base can be used for the logarithms as long as both sides use the same one; base 10 is the calculator's $\log$ button. A solution exists only when $b>0$, because $a^x$ is always positive. Example: $5^x=40$ gives $x=\frac{\log40}{\log5}=2.29$ (3 s.f.). Exactly, $x=\log_540$.

The change of base formula lets you evaluate a logarithm with any base on a calculator: $\log_ab=\frac{\log_cb}{\log_ca}.$ Proof: let $y=\log_ab$, so $a^y=b$. Taking $\log_c$ of both sides, $y\log_ca=\log_cb$, which gives the formula. So the solution of $a^x=b$ is $x=\log_ab=\frac{\log_{10}b}{\log_{10}a}$. For example $\log_530=\frac{\log_{10}30}{\log_{10}5}=2.11$.

When the index is an expression such as $2x+1$, either isolate the power first or take logarithms straight away. For $3^{2x+1}=600$:

• Method 1: $3\times3^{2x}=600$, so $3^{2x}=200$, $2x=\frac{\log200}{\log3}=4.82\ldots$, $x=2.41$.
• Method 2: $(2x+1)\log3=\log600$, so $2x+1=\frac{\log600}{\log3}=5.82\ldots$ and $x=2.41$. Keep full calculator values until the end and round only the final answer, to 3 significant figures unless told otherwise.

If both sides are powers with different bases, take logarithms of both sides, then collect the terms in $x$ and factorise. For $7^x=2^{x+3}$: $x\log7=(x+3)\log2\;\Rightarrow\;x(\log7-\log2)=3\log2\;\Rightarrow\;x=\frac{3\log2}{\log7-\log2}=1.66.$ The change of base formula also links answers: since $9^x=100$ gives $x=\frac{\log100}{\log9}=\frac2{2\log3}=\frac1{\log3}=\log_310=2.10$.

Compound interest follows $V=P\times r^n$, where $r$ is the yearly multiplier. To find when an amount is reached, set up $a^x=b$. For $2000\times1.04^n=3000$: $1.04^n=1.5$, so $n=\frac{\log1.5}{\log1.04}=10.3$ years. Interpret the answer in context: the account first passes £3000 during the 11th year, so after 11 complete years its value is $2000\times1.04^{11}=\text{£}3079$. To compare two accounts, equate them: $2000\times1.04^n=1500\times1.06^n$ gives $\left(\frac{1.06}{1.04}\right)^n=\frac43$ and $n=15.1$.