# How to solve this logarithmic equation whose expressions have different bases?

I have been trying to solve the following equation for a while and i can't seem to figure it out, your help would be greatly appreciated. Here is the equation: $3^x$=$5^{x-1}$

• If $x$ is an integer, there is no solution. If $a$and $b$ are coprime, then $a^m=b^n$ has no solutions for integer $n$ and $m$ – Joao Oct 9 '14 at 2:29
• Take the logarithm of both sides, any base you like. I would use the natural logarithm (base $e$), but if you prefer $10$, that's fine too. – André Nicolas Oct 9 '14 at 2:31

Hints:

$$5^{x-1}=\frac {5^x}5\\ \frac{a^q}{b^q}=\left(\frac ab\right)^q$$

• I had considered this but did not think it was the right thing to do. Thanks for pointing me in the right direction. – FutureSci Oct 9 '14 at 2:47

Hint: $$\log_3(5^{x-1})=(x-1)\log_3(5)$$

• Or, perhaps more practically, $\ln(5^{x-1})=(x-1)\ln 5$. +1 for you either way. – MPW Oct 9 '14 at 2:37

Here are the steps $$3^x=5^{x-1}$$ $$1=\frac{5^{x-1}}{3^x}= 5^{x-1}3^{-x}$$ $$1= e^{\ln(5^{x-1}3^{-x})} = e^{\ln(5^{x-1})+\ln(3^{-x})}$$ $$\ln(1)= \ln(5^{x-1})+\ln(3^{-x})$$ $$0= (x-1)\ln(5)-x\ln(3)$$ $$\ln(5)= x\ln(5)-x\ln(3)$$ $$\ln(5)= x(\ln(5)-\ln(3))$$ $$x=\frac{\ln(5)}{\ln(5)-\ln(3)}$$

$$(\frac53)^x = 5$$ $$x = \frac{\ln 5}{\ln \frac53} =...$$