JAMB 2004 · UME · Q8

If $6\log_x 2 - 3\log_x 3 = 3\log_5 0.2$ , find $x$ .

Worked solution (try it first)

Right side: $0.2 = 5^{-1}$ , so $3\log_5 0.2 = 3(-1) = -3$ .
Left side: $6\log_x 2 = \log_x 64$ and $3\log_x 3 = \log_x 27$ , so the left side is $\log_x \frac{64}{27}$ .
Change to index form: $x^{-3} = \frac{64}{27}$ , so $x^3 = \frac{27}{64}$ .
Take cube roots: $x = \frac34$ , option B.