SOLUTION: I'm having trouble solving for 'x' in this equation {{{log4(x-2)+ log4(x+6)=2}}}. I have tried re-arranging it to make an exponential equation, but I end up confusing myself even

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Question 98545This question is from textbook Calculus with Applications Brief Version
: I'm having trouble solving for 'x' in this equation log4%28x-2%29%2B+log4%28x%2B6%29=2. I have tried re-arranging it to make an exponential equation, but I end up confusing myself even more.
Thank you for looking at this. This question is from textbook Calculus with Applications Brief Version

Found 2 solutions by stanbon, bimanewton:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log4(x-2)+ log4(x+6)=2
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Note: Keep the "base 4" in mind
============
log(x-2)(x+6) = 2
(x-2)(x+6) = 4^2
x^2+4x-12 = 16
x^2+4x-28=0
x = [-4 +- sqrt(16-4*-28]/2
x = [-4 +- sqrt(128)]/2
----------------
x = [-4 + 8sqrt(2)]/2 = -2+4sqrt2
OR
x = [-4 - 8sqrt(2)]/2 = -2-4sqrt2
====================
Cheers,
Stan H.

Answer by bimanewton(8) About Me  (Show Source):