# Logarithmic equation

• Oct 15th 2010, 07:41 AM
regdude
Logarithmic equation
Hi!
Today at math class we were given some examples to solve, all were easy to solve, except for this one:
http://img257.imageshack.us/img257/8409/logr.jpg
I tried to rewrite 3^log(3,x) as simply x and (3^log(2,x))^2 as 3^2log(2,x) or 3^log(2,x^2), but couldn't see anything to do further.
I asked the teacher and she didn't seem to know either :P
• Oct 15th 2010, 08:07 AM
earboth
Quote:

Originally Posted by regdude
Hi!
Today at math class we were given some examples to solve, all were easy to solve, except for this one:
http://img257.imageshack.us/img257/8409/logr.jpg
I tried to rewrite 3^log(3,x) as simply x and (3^log(2,x))^2 as 3^2log(2,x) or 3^log(2,x^2), but couldn't see anything to do further.
I asked the teacher and she didn't seem to know either :P

That must be a typo and the equation should read:

$\left(3^{\log_2 x} \right)^2 - 4 \cdot 3^{\log_2 x} + 3 = 0$

Substitute $y = 3^{\log_2 x}$

and solve for y.

Consequently re-substitute and solve for x.

I've got $x = 2~\vee~x=1$

If there isn't a typo you can only use numerical methods (for instance Newton) to get an approximate solution.
• Oct 15th 2010, 08:14 AM
regdude
Well not a typo in my notebook, the book has exactly how I had it.
I tried to use Microsoft Math to solve it for me, aprox. answer was 1.
The answer in book was x=2 and x=1.
So to get to the answer the second logarithm base must be 2.
Thanks for clearing that up :)