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SolidSnake
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#1
Report Thread starter 14 years ago
#1
Hey everyone
I need some help on logs. I know how to solve most of them, but how do I solve something like this?:

3^2x = 4^(2-x)

Thanks
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Gauss
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#2
Report 14 years ago
#2
(Original post by SolidSnake)
Hey everyone
I need some help on logs. I know how to solve most of them, but how do I solve something like this?:

3^2x = 4^(2-x)

Thanks
Take logs of both sides:

log3^2x = log4^(2-x)
=> 2xlog3 = (2-x)log4
=> 2xlog3 = 2log4 - xlog4
=> x(2log3 + log4) = 2log4
=> x = 2log4/(2log3 + log4)
=> x = log16/(log9 + log4)
=> x = log16/log36
=> x = 2log4/2log6
=> x = log4/log6

Euclid
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kikzen
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#3
Report 14 years ago
#3
take logs on both side

ln 3^2x = ln 4^2-x
then ... by log laws,
2x ln3 = (2-x)ln4
rearrange to find x, remebering that ln3 and ln4 are just numbers
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SolidSnake
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#4
Report Thread starter 14 years ago
#4
Thank You
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