Solve for $x$: $2^{x} = \dfrac{1}{32}$.

Solve for $x$: $2^{x} = \dfrac{1}{32}$.
  1. $5$
  2. $-5$
  3. $\dfrac{1}{5}$
  4. $-\dfrac{1}{5}$

Solution

$\dfrac{1}{32} = \dfrac{1}{2^{5}} = 2^{-5}$. So $x = -5$.

Asked in: IMO

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