2log(base4)X - log(base4)(x+3)=1
Note that $\displaystyle 2\log_4x-\log_4\left(x+3\right)=1\implies \log_4\left(x^2\right)-\log_4\left(x+3\right)=1$
Now recall that $\displaystyle \log_ab-\log_ac=\log_a\left(\frac bc\right)$
Thus, $\displaystyle \log_4\left(x^2\right)-\log_4\left(x+3\right)=1\implies\log_4\left(\frac{ x^2}{x+3}\right)=1$
Now, note that $\displaystyle \log_44=1$
Thus, we now have the equation $\displaystyle \log_4\left(\frac{x^2}{x+3}\right)=\log_44$
Since we want the two logarithms to be equal, we must see where $\displaystyle \frac{x^2}{x+3}=4$
Can you continue on from here?
--Chris