Logarithmic equation

**cristinaivelisse** · August 20th 2010, 05:36 PM

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

**Ackbeet** · August 20th 2010, 06:03 PM

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?

**cristinaivelisse** · August 20th 2010, 06:25 PM

I'm sorry

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?

**skeeter** · August 20th 2010, 06:35 PM

if $\log_b(a) = c$ , then $b^c = a$

change your log equation to exponential form.

**Educated** · August 20th 2010, 11:07 PM

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...

Thread: Logarithmic equation

LinkBack

Thread Tools

Display