Solve $\left(\dfrac{2}{3}\right)^{x-1} = \dfrac{27}{8}$.
Solve $\left(\dfrac{2}{3}\right)^{x-1} = \dfrac{27}{8}$.
- $-2$
- $-3$
- $2$
- $4$
Solution
$\dfrac{27}{8} = \left(\dfrac{3}{2}\right)^{3} = \left(\dfrac{2}{3}\right)^{-3}$. So $x - 1 = -3 \Rightarrow x = -2$.
Asked in: IMO
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