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Question Number 56336 by otchereabdullai@gmail.com last updated on 14/Mar/19

$$\mathrm{make}\mathrm{r}\mathrm{the}\mathrm{subject}\mathrm{of}$$$$\mathrm{P}={(1+\frac{\mathrm{r}}{100})}^{\mathrm{t}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Mar/19

$$questionnotclear....$$$$p={(1+\frac{r}{100})}^t$$$$p^{\frac{1}{t}}=1+\frac{r}{100}$$$$r=100(p^{\frac{1}{t}}-1)$$

Commented by otchereabdullai@gmail.com last updated on 14/Mar/19

$$\mathrm{thank}\mathrm{you}\mathrm{prof}\mathrm{am}\mathrm{much}\mathrm{greatful}$$

Commented by otchereabdullai@gmail.com last updated on 14/Mar/19

$$\mathrm{prof}\mathrm{how}\mathrm{did}\mathrm{get}\mathrm{the}{\mathrm{p}}^{\frac{1}{\mathrm{t}}}\mathrm{i}\mathrm{want}\mathrm{to}$$$$\mathrm{understand}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Mar/19

$$supose$$$$x^a=b$$$$nowtoeliminate\mathit{a}fromthepowerofx$$$${\mathit{x}}^{\mathit{a}}=\mathit{b}$$$${\left({\mathit{x}}^{\mathit{a}}\right)}^{\frac{1}{\mathit{a}}}={\left(\mathit{b}\right)}^{\frac{1}{\mathit{a}}}$$$${\mathit{x}}^{\mathit{a}\times \frac{1}{\mathit{a}}}={\mathit{b}}^{\frac{1}{\mathit{a}}}$$$$\mathit{x}={\mathit{b}}^{\frac{1}{\mathit{a}}}$$

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