How do you solve $4^x = 7^(x-4)$ ?

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Answer 1 · 2015-12-04T22:08:08

$x~= -6.7745$

Given the exponential equation $4^x = 7^(x-4)$

To solve exponential equation we can use logarithm.

Step 1: Take log of both side

$log 4^x = log 7^(x-4)$

Using the power rule of logarithm

$x log 4 = (x-4) log 7$

Then distribute

$x log 4 = x log 7 - 4 log 7$

Then bring all the "x" on one side

$x log 4 - x log 7 = -4 log 7$

Factor out the greatest common factor

$x(log 4 - log 7) = -4 log 7$

Isolate "x"

$x = (-4log 7)/(log 4 - log 7)$

$x~= -6.7745$

How do you solve 4^x = 7^(x-4)4^x = 7^(x-4)?