1. ## Logarithmic and radical equations

Greetings,
I have encountered inability to resolve the below shown problem.

2. Hi

$\displaystyle 2\:\log_{4}(3y-4) - \log_{2}(x-4) = 1 \implies\:$$\displaystyle \:2\:\frac{\ln(3y-4)}{\ln 4} -\frac{\ln(x-4)}{\ln 2} = 1$

and since ln(4) = ln(2²) = 2 ln(2)

$\displaystyle \ln(3y-4)-\ln(x-4) = \ln 2 \implies \ln \frac{3y-4}{x-4} = \ln 2 \implies \frac{3y-4}{x-4} = 2 \implies x = \frac32 \:y + 2$

Now you can substitute this expression into the first equation