How do you solve $9^{8 x} = 13^{- x + 7}$ $9^(8x)=13^(-x+7)$ ?

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Answer 1 · 2016-02-27T23:36:49

$x = 7 - 8 \frac{log (9)}{log (13)}$ $x=7-8log(9)/log(13)$

$9^{8 x} = 13^{- x + 7}$ $9^(8x)=13^(-x+7)$

You need to use logarithms:

$8 x . log (9) = (- x + 7) log (13)$ $8x. log(9)=(-x+7)log(13)$

$8 \frac{log (9)}{log (13)} = - x + 7$ $8log(9)/log(13)=-x+7$

$x = 7 - 8 \frac{log (9)}{log (13)}$ $x=7-8log(9)/log(13)$

It is a very ugly solution.

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