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

Question

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: $3^x$=$5^{x-1}$

Answer 1

Hints:

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

Answer 2

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

Answer 3

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

Answer 4

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