# Logarithmic equation

• Aug 20th 2010, 06:36 PM
cristinaivelisse
Logarithmic equation
log(base)3 (5x-1)=2 I need help solving this equation. I hope is clear. It's suppose to be log base 3 and 5x-1 is whole term :) thank you for your attention
• Aug 20th 2010, 07:03 PM
Ackbeet
I would say the first thing to do is raise 3 to both sides of the equation thus:

$\displaystyle{3^{\log_{3}(5x-1)}=3^{2}}.$

Can you see where to go from here?
• Aug 20th 2010, 07:25 PM
cristinaivelisse
I'm sorry (Crying) I have no idea were to go from there. Do I move the log base 3 (5x-1) up front the 3... like:

3 log base 3 (5x-1) and then distribute?
• Aug 20th 2010, 07:35 PM
skeeter
if $\log_b(a) = c$ , then $b^c = a$

change your log equation to exponential form.
• Aug 21st 2010, 12:07 AM
Educated
Quote:

Originally Posted by Ackbeet
$\displaystyle{3^{\log_{3}(5x-1)}=3^{2}}.$

$\displaystyle{3^{\log_{3}(5x-1)}=3^{2}}$

$\displaystyle{3^{\log_{3}(a) = a}$

Therefore the ${3^{\log_{3}}$ cancels itself out, leaving:

$\displaystyle{(5x-1)=3^{2}}$

You know where to go from here...