Question
Make $R$ the subject of formula $A=P\left(1+\frac{R}{100}\right)^N$
✓
Answer
$A=P\left(1+\frac{R}{100}\right)^N $
$\Rightarrow \frac{A}{P}=\left(1+\frac{R}{100}\right)^N$
Taking $N$ the root both sides
$\Rightarrow\left(\frac{A}{P}\right)^{\frac{1}{N}}=\left(1+\frac{R}{100}\right) $
$\Rightarrow\left(\frac{A}{P}\right)^{\frac{1}{N}}-1=\frac{R}{100} $
$\Rightarrow 100\left(\left(\frac{A}{P}\right)^{\frac{1}{N}}-1\right)=R $
$\Rightarrow R=100\left(\sqrt[N]{\frac{A}{P}}-1\right) .$
$\Rightarrow \frac{A}{P}=\left(1+\frac{R}{100}\right)^N$
Taking $N$ the root both sides
$\Rightarrow\left(\frac{A}{P}\right)^{\frac{1}{N}}=\left(1+\frac{R}{100}\right) $
$\Rightarrow\left(\frac{A}{P}\right)^{\frac{1}{N}}-1=\frac{R}{100} $
$\Rightarrow 100\left(\left(\frac{A}{P}\right)^{\frac{1}{N}}-1\right)=R $
$\Rightarrow R=100\left(\sqrt[N]{\frac{A}{P}}-1\right) .$
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