# Math

Solve for the inequality:

1/3^x (1/3^x - 2) < 15

I know the answer is x > - (log 5/log 3), can someone explain me how?

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1. suppose we just look at
1/3^x (1/3^x - 2) = 15
3^-x(3^-x - 2) = 15
let 3^-x = a for easier typing
then we have
a(a-2) = 15
a^2 - 2a - 15 = 0
(a-5)(a+3) = 0
a = 5 or a = -3

3^-x = 5
log 3^-x = log5
-x log3 = log5
-x = log5/log3
x = -log5/log3

the other part of 3^-x = -3 has no solution

but we wanted
1/3^x (1/3^x - 2) < 15
so x < -log5/log3

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posted by Reiny

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