2 How do I solve this equation involving a logarithm? 2 How do I solve this equation involving a logarithm? 3
$$\log_2\left(\frac{x}{2}\right) = \log_3\left(\frac{2+x}{3}\right)$$

$$\log_2(x)-\log_2(2) = \log_3(2+x)-\log_3(3)$$

$$\log_2(x)-1 = \log_3(2+x)-1$$

$$\log_2(x) = \log_3(2+x)$$

$$\log_3(2+x)=\frac{\log_{10}(2+x)}{\log_{10}(3) }$$

$$\log_2(x) = \frac{\log_{10}(x)}{\log_{10}(2)}$$

$$\therefore \frac{\log_{10}(2+x)}{\log_{10}(3)} = \frac{\log_{10}(x)}{\log_{10}(2)}$$
From now on, we'll write $\log_{10}(a)$ as $\log(a)$.

$$\log(2+x)\log(2) = \log(3)\log(x)$$

$$\log(2+x) = \frac{\log(3)}{\log(2)}\log(x)$$

$$ \frac{\log(3)}{\log(2)} \approx 1.585\ldots$$

$$\log(2+x) = 1.585\log(x)$$

$$\log(2+x) = \log(x^{1.585})$$

I don't have the time to finish off the answer right now, I'll do it later. Hopefully it should be obvious, just subtract and multiply out.
Use the properties of the log function: 

$\log_a {\frac{b}{c}} = \log_a b -\log_a c$ 

and $\log_a x = \log_a b \times \log_b x$.