How do you solve $4^{x - 3} = \frac{1}{16}$ $4^(x-3)=1/16$ ?

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Answer 1 · 2016-09-10T07:21:51

$x = 1$ $x =1$

There are 2 approaches which I use for exponential equations, which are equations which have indices.

$\to$ $rarr$ make the bases the same
$\to$ $rarr$ make the indices the same

If neither of these work, then use logs.

We should know that 16 is a power of 4, so use 4 as the base on both sides.

$4^{x - 3} = \frac{1}{16} = \frac{1}{4^{2}}$ $4^(x-3) = 1/16 = 1/4^2$

$4^{x - 3} = 4^{- 2} \leftarrow$ $4^(x-3) = 4^-2 " "larr$ now equate the indices

$x - 3 = - 2$ $x-3 = -2$

$x = - 2 + 3 = 1$ $x = -2+3 = 1$

How do you solve 4x−3=1164^(x-3)=1/16?