SOLUTION: how do you solve for x in the following equation:
2^(2x+12) = 3^(x-45)
i changed it to log2 (2x+12) = log3 (x-45)
is that right? where do i go from there?
Question 397898: how do you solve for x in the following equation:
2^(2x+12) = 3^(x-45)
i changed it to log2 (2x+12) = log3 (x-45)
is that right? where do i go from there? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! 2^(2x+12) = 3^(x-45)
i changed it to log2 (2x+12) = log3 (x-45)
is that right? where do i go from there?
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(2x+12)*log(2) = (x-45)*log(3) I think that's what you meant anyway.
Just solve for x. Logs of constants are constants.
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2x*log(2) + 12log(2) = x*log(3) - 45log(3)
2x*log(2) - x*log(3) = -12log(2) - 45log(3)
x*(log(3) - 2log(2)) = 12log(2) + 45log(3)
x = (12log(2) + 45log(3))/(log(3) - 2log(2))
x =~ -200.7609
