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#