JAMB 1983 · UME · Q32

If $\left(\frac23\right)^m\left(\frac34\right)^n = \frac{256}{729}$ , find the values of $m$ and $n$ .

Worked solution (try it first)

Split each fraction into primes: $\left(\frac23\right)^m = 2^m \times 3^{-m}$ and $\left(\frac34\right)^n = 3^n \times 2^{-2n}$ , because $4 = 2^2$ .
Collect the powers: the left side is $2^{m - 2n} \times 3^{n - m}$ .
The right side is $\frac{256}{729} = 2^8 \times 3^{-6}$ .
Match the powers: $m - 2n = 8$ and $n - m = -6$ .
Add the two equations: $-n = 2$ , so $n = -2$ .
Then $m = 8 + 2n = 8 - 4 = 4$ .
So $m = 4$ , $n = -2$ , option D.