How do you solve $3^{x-1}$ = $2^x$ using natural logs

Question

My pre-calc book wants me to solve $3^{x-1} = 2^x$ using natural logs. I get (x-1)ln(3) = xln(2). But from there I don't know where to go. The book answer is $\frac{ln(3)}{ln(3) - ln(2)}$ can someone please expain to me the steps to get there?

Answer 1

Just isolate $x$ to get $x\ln3 - \ln3 = x\ln 2 \implies x(\ln3 - \ln 2) = \ln 3$

$$x =\frac{\ln 3}{\ln3 -\ln2}$$

Answer 2

Hint:$$3^{x-1}=2^x\iff\left(\frac32\right)^x=3$$

Answer 3

$$x \ln 3 -\ln 3 = x\ln 2$$

so if we swap $x\ln2 $ and $\ln 3$ we get $$x \ln 3 -x\ln 2 = \ln 3$$

so $$x (\ln 3 -\ln 2) = \ln 3$$

so $$x = {\ln 3\over \ln 3 -\ln 2}$$

Answer 4

$x\ln(3)-\ln(3)=x\ln(2) \implies x(\ln(3)-\ln(2))=\ln(3)$ hence $x=\frac{\ln(3)}{\ln(3)-\ln(2)}$

Answer 5

$$(x-1)\ln3=x\ln 2$$ $$x(\ln 3-\ln 2)=\ln 3$$ $$x=\frac {\ln 3}{\ln 3-\ln 2}$$