Solving log equation

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September 4th 2010, 05:19 AM
Garas

Solving log equation

How to find x in this equation: http://image.beyluxe.com/pictures/2d...07ae434cb7.jpg
September 4th 2010, 06:05 AM
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Quote:

Originally Posted by Garas

How to find x in this equation: http://image.beyluxe.com/pictures/2d...07ae434cb7.jpg

There is no solution.

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

But $\log_5(1-3^{\log_{15}(2/9)}) \ne \log_{15}(2/9)$
September 4th 2010, 11:40 AM
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 "+"?
September 4th 2010, 01:24 PM
Garas

no, that math problem was in a test for the entrance to the college

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