make-r-the-subject-of-P-1-r-100-t-

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

make r the subject of

P = {(1 + \frac{r}{100})}^{t}

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

q u e s t i o n n o t c l e a r \dots .

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

thank you prof am much greatful

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

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

understand

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

s u p o s e

x^{a} = b

n o w t o e l i m i n a t e a f r o m t h e p o w e r o f x

x^{a} = b

{(x^{a})}^{\frac{1}{a}} = {(b)}^{\frac{1}{a}}

x^{a \times \frac{1}{a}} = b^{\frac{1}{a}}

x = b^{\frac{1}{a}}