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Question Number 8663 by uchechukwu okorie favour last updated on 20/Oct/16

if I=a(1−(r/(100)))^n   make n the subject of formular

$${if}\:{I}={a}\left(\mathrm{1}−\frac{{r}}{\mathrm{100}}\right)^{{n}} \\ $$$${make}\:{n}\:{the}\:{subject}\:{of}\:{formular} \\ $$$$ \\ $$

Answered by sandy_suhendra last updated on 20/Oct/16

(1−(r/(100)))^n =(I/a)  log (1−(r/(100)))^n = log (I/a)  n log(1−(r/(100))) = log (I/a)  n = ((log(I/a))/(log (1−(r/(100)))))  n = log_((1−(r/(100)))) [(I/a)]

$$\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{100}}\right)^{\mathrm{n}} =\frac{\mathrm{I}}{\mathrm{a}} \\ $$$$\mathrm{log}\:\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{100}}\right)^{\mathrm{n}} =\:\mathrm{log}\:\frac{\mathrm{I}}{\mathrm{a}} \\ $$$$\mathrm{n}\:\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{100}}\right)\:=\:\mathrm{log}\:\frac{\mathrm{I}}{\mathrm{a}} \\ $$$$\mathrm{n}\:=\:\frac{\mathrm{log}\frac{\mathrm{I}}{\mathrm{a}}}{\mathrm{log}\:\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{100}}\right)} \\ $$$$\mathrm{n}\:=\:\mathrm{log}_{\left(\mathrm{1}−\frac{\mathrm{r}}{\mathrm{100}}\right)} \left[\frac{\mathrm{I}}{\mathrm{a}}\right] \\ $$

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