Question

Given 4725 = $3^a{5}^b{7}^c$, find:

(i) the integral values of a, b and c

(ii) the value of 2${}^{-a}{3}^b{7}^c$

Answer 1

4725 = $3^a{5}^b{7}^c$

(i) $4725=3^3.5^2.7^1$ = 3${}^a.5^b.7^c$ Comparing, we get a = 3, b = 2 and c = 1.

(ii) 2${}^{-a}{3}^b{7}^c$ = 2${}^{-3}\times 3^2\times 7^1$ = $\frac{1}{2^3}\times 3^2\times 7^1$ = $\frac{3\times 3\times 7}{2\times 2\times 2}$ = $\frac{63}{8}$