Finding $\frac{3x+y}{6x-1} = ?$ $$2^{3x} = 18$$
$$2^y = 9$$
$$\frac{3x+y}{6x-1} = ?$$
Let me show my attempt:
$$2^y = 9, 2^y = 3^2, y = 1$$
$$2^{3x} = 18, x = 1$$
I think I've gone too wrong

You can get the value of $x$ by doing :
$2^{3x} = 18 \implies \ln(2^{3x})=\ln(18) \implies 3x \cdot \ln(2)= \ln(18) \implies 3x = \frac{\ln(18)}{\ln(2)} \implies x=\frac{\ln(18)}{3 \cdot \ln(2)}$
And similarly, the value of $y$ by doing:
$2^y = 9 \implies \ln(2^y) = \ln(9) \implies y \cdot \ln(2) = \ln(9) \implies y= \frac{\ln(9)}{\ln(2)}$
And then simply substitute
$$\frac{3x+y}{6x-1} = \frac{3 \cdot\frac{\ln(18)}{3 \cdot \ln(2)}+\frac{\ln(9)}{\ln(2)}}{6 \cdot \frac{\ln(18)}{3 \cdot \ln(2)}-1}=\frac{\frac{\ln(18)+\ln(9)}{\ln(2)}}{\frac{2 \ln(18)-\ln(2)}{\ln(2)}} = \frac{\ln(18) + \ln(9)}{2\cdot \ln(18) - \ln(2)} = \frac{\ln(18 \cdot 9)}{\ln(\frac{18^2}{2})} = \frac{\ln(162)}{\ln(162)} = 1.$$
