How do you solve $4^{x - 3} = 32$ $4^ { x - 3} = 32$ ?

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Answer 1 · 2016-11-01T19:02:55

$x = \frac{11}{2} = 5 \frac{1}{2} = 5.5$ $x = 11/2 = 5 1/2 = 5.5$

When you have a variable in the index, you are working with an exponential equation.

There are 2 types of exponential equations

This example is an example of the former, because both bases are powers of 2.

$4^{x - 3} = 32$ $4^(x-3) = 32$

${(2^{2})}^{x - 3} = 2^{5}$ $(2^2)^(x-3) = 2^5$

$2^{2 x - 6} = 2^{5} \leftarrow$ $2^(2x-6) = 2^5" "larr$ if the bases are equal, the indices are equal

$:. 2x-6 = 5$

$2x = 5+6$

$2x =11$

$x = 11/2 = 5 1/2 = 5.5$

How do you solve 4^ { x - 3} = 324x−3=324^ { x - 3} = 32?