Solve $3^{x+1} = 2^x$

Question

So I have solved it two different ways, using $\log_2 $ of both sides:
$x=x+1(\log_2 3)$
$x=x \log_2 3 + \log_2 3$
$ \log_2 3 = x(1- \log_2 3)$
$x= \frac {\log_2 3} {1-\log_2 3}$
$x= -2.71$
And $\log_3$ of both sides:
$x+1=x(\log_3 2)$
$1=x(\log_3 2 -1)$
$x= \frac {1} {\log_3 2 -1}$
$x=-2.71$
However, is it also possible to use change of base law with the $\log_3$ version?
$x+1=\frac {\log_2 (2^x)} {\log_2 3}$
How would you simplify $\log_2 3 = \frac {x} {x+1}$ ?

Answer 1

Hint $$3^{x+1}=2^x\iff \left(\frac{2}{3}\right)^x=3.$$

Answer 2

HINT

We can simply use natural log to obtain

$$3^{x+1}= 2^x\iff (x+1)\log 3=x\log 2$$

then find $x$.

Answer 3

$$3^{x+1} = 2^x \implies \sqrt[x]{3^{x+1}} = 2 \implies 3^{\frac{x +1}{x}} = 2$$ $$\left(\frac{x +1}{x}\right) = \log_32$$ $$\left(1+\frac{1}{x}\right) = \log_32$$ $$x = \frac{1}{\left(\log_32-1\right)}$$ $$x = -\frac{\ln \left(3\right)}{\ln \left(\frac{3}{2}\right)}$$