# Logarithmic equation

• Aug 20th 2010, 05:36 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, 06:03 PM
Ackbeet
I would say the first thing to do is raise 3 to both sides of the equation thus:

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

Can you see where to go from here?
• Aug 20th 2010, 06:25 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, 06:35 PM
skeeter
if \$\displaystyle \log_b(a) = c\$ , then \$\displaystyle b^c = a\$

change your log equation to exponential form.
• Aug 20th 2010, 11:07 PM
Educated
Quote:

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

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

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

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

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

You know where to go from here...