1. ## log problem

I have a log problem $\log_{3}x-2\log_{x}3=1$ the question is asking to find the values of x.

I started the problem by changing the base

$\log_{3}x-\frac {2}{\log_{3}x}-1=0$

$(\log_{3}x)^2-2-\log_{3}x=0$

$y^2-y-2=0$

Am I on the right track? Can someone help me with this problem?

2. Originally Posted by scrible
I have a log problem $\log_{3}x-2\log_{x}3=1$ the question is asking to find the values of x.

I started the problem by changing the base

$\log_{3}x-\frac {2}{\log_{3}x}-1=0$

$(\log_{3}x)^2-2-\log_{3}x=0$

$y^2-y-2=0$

Am I on the right track? Can someone help me with this problem?
factor and finish ... don't forget to check your solutions for validity in the original equation.

3. Originally Posted by skeeter
factor and finish ... don't forget to check your solutions for validity in the original equation.

When I use the formula I am not getting the answers that are in the book. which is $9,\frac{1}{3}$

4. Originally Posted by scrible
When I use the formula I am not getting the answers that are in the book. which is $9,\frac{1}{3}$
$(\log_3{x})^2 - \log_3{x} - 2 = 0
$

$(\log_3{x} - 2)(\log_3{x} + 1) = 0$

$\log_3{x} = 2$ ... $x = 9$

$\log_3{x} = -1$ ... $x = \frac{1}{3}$

5. Originally Posted by skeeter
$(\log_3{x})^2 - \log_3{x} - 2 = 0
$

$(\log_3{x} - 2)(\log_3{x} + 1) = 0$

$\log_3{x} = 2$ ... $x = 9$

$\log_3{x} = -1$ ... $x = \frac{1}{3}$
thanks I got it now. It is the last part of these problem that gives me trouble.