How do you solve the exponential equation $5^{3 x} = 25^{x - 4}$ $5^(3x)=25^(x-4)$ ?

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Answer 1 · 2017-08-24T06:23:34

$x = - 8$ $x=-8$

As a first approach to solving exponential equations, try to either:

In this case, notice that $25$ $25$ is a power of $5, 25 = 5^{2}$ $5," "color(blue)(25 =5^2)$

$5^{3 x} = 25^{x - 4}$ $5^(3x) = color(blue)(25)^(x-4)$

$5^{3 x} = {(5^{2})}^{x - 4} \leftarrow$ $5^(3x) = color(blue)((5^2))^(x-4)" "larr$ multiply the indices

$5^{3 x} = 5^{2 x - 8} \leftarrow$ $color(red)(5)^(3x) = color(red)(5)^(2x-8)" "larr$ the bases are the same

$:. 3x = 2x-8" "larr$ the indices are equal

$3x -2x =-8$

$x=-8$

How do you solve the exponential equation 5^(3x)=25^(x-4)53x=25x−45^(3x)=25^(x-4)?