Where to go from here?
$\displaystyle \displaystyle \begin{align*} \frac{x + 1}{x} &= \frac{\log{5}}{\log{4}} \\ 1 + \frac{1}{x} &= \frac{\log{5}}{\log{4}} \\ \frac{1}{x} &= \frac{\log{5}}{\log{4}} - 1 \\ \frac{1}{x} &= \frac{\log{5} - \log{4}}{\log{4}} \\ x &= \frac{\log{4}}{\log{5} - \log{4}} \end{align*}$
Your calculations are OK, but there is a simpler way to do this question:
$\displaystyle \displaystyle{5^x=4^{x+1}~\implies~5^x=4 \cdot 4^{x}~\implies~\left(\frac54 \right)^x=4~\implies~x = \log_{\frac54}(4)}$
With your result:
$\displaystyle \displaystyle{(x)log5=(x+1)log4~\implies~(x)log5=( x)log4 + log4~\implies~(x)log5-(x)log4 = log4~\implies~x=\frac{log4}{log5-log4}}$
$\displaystyle \frac{3x+1}{x-2} = \frac{\log{3}}{\log{2}}$
cross multiply ...
$\displaystyle (3x+1)\log{2} = (x-2)\log{3}$
distribute both sides ...
$\displaystyle 3x\log{2} + \log{2} = x\log{3} - 2\log{3}$
get all x terms on the same side ...
$\displaystyle 3x\log{2} - x\log{3} = -\log{2} - 2\log{3}$
factor out the x ...
$\displaystyle x(3\log{2} - \log{3}) = -(\log{2}+2\log{3})$
$\displaystyle x\log\left(\frac{8}{3}\right) = -\log{18}$
solve for x ...
$\displaystyle x = -\frac{\log{18}}{\log\left(\frac{8}{3}\right)}$
No, when you have x in the denominator, you need to reciprocate. It should be like this...
$\displaystyle \displaystyle \begin{align*} 3 + \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} \\ \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} - 3 \\ \frac{7}{x - 2} &= \frac{\log{3} - 3\log{2}}{\log{2}} \\ \frac{x - 2}{7} &= \frac{\log{2}}{\log{3} - 3\log{2}} \\ x - 2 &= \frac{7\log{2}}{\log{3} - 3\log{2}} \\ x &= \frac{7\log{2}}{\log{3} - 3\log{2}} + 2 \\ x &= \frac{\log{2} + 2\log{3}}{\log{3} - 3\log{2}} \end{align*}$