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
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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?
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?
if $\displaystyle \log_b(a) = c$ , then $\displaystyle b^c = a$ change your log equation to exponential form.
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...
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