# Solving log equation

• Sep 4th 2010, 06:19 AM
Garas
Solving log equation
• Sep 4th 2010, 07:05 AM
undefined
Quote:

Originally Posted by Garas

There is no solution.

$\log_{15}(2/9)=\log_3x\implies x=3^{\log_{15}(2/9)}$

But $\log_5(1-3^{\log_{15}(2/9)}) \ne \log_{15}(2/9)$
• Sep 4th 2010, 12:40 PM
HallsofIvy
Those are two separate equations, $log_{15}\frac{2}{9}= log_3 x$ and $log_3 x= log_5(1- x)$. As undefined said, there is no value of x that satifies both. Was one of those "=" supposed to be "-" or "+"?
• Sep 4th 2010, 02:24 PM
Garas
no, that math problem was in a test for the entrance to the college