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

(i) Given $4725=3^a{5}^b{7}^c$
First find out the prime factorisation of 4725.
$\begin{array}{cc}3& 4725\\ 3& 1575\\ 3& 525\\ 5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}$
It can be observed that 4725 can be written as $3^3\times 5^2\times 7^1$.
$\therefore 4725=3^a{5}^b{7}^c=3^3{5}^2{7}^1$
Hence, a = 3, b = 2 and c = 1.

(ii)
When a = 3, b = 2 and c = 1,
$$\begin{gathered} 2^{-a}{3}^b{7}^c \\ =2^{-3}\times 3^2\times 7^1 \\ =\frac{1}{8}\times 9\times 7 \\ =\frac{63}{8} \end{gathered}$$
Hence, the value of $2^{-a}{3}^b{7}^c$ is $\frac{63}{8}$.