Logarithms 102

$\begin{aligned} \log_2 x + \log_2 6 & = 3 \\ \log_2 (6x) & = \log_2 (2^3) \\ 6x & = 8 \\ \implies x & = \frac 86 \approx \boxed{1.33} \end{aligned}$

Using logarithm properties, we can combine the 2 logs on the left hand side: $\log_{2}x + \log_{2}6 = 3 \Rightarrow \log_{2}6x = 3$

Now we can raise everything to a power of 2 and cancel out the log: $\log_{2}6x = 3 \Rightarrow 2^{\log_{2}6x} = 2^{3} \Rightarrow 6x = 8$

Now using simple rearranging we get: $\huge 6x = 8 \Rightarrow x=\boxed{\frac{4}{3}}$