Determine if the equation is valid/true The equation is:
$$\log_b \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = 2\log_b(\sqrt{3}+\sqrt{2}).$$
I can get as far as:
$$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$$
Which looks almost too simple, but I can't get the signs to match up right to solve the problem.  Do I need to further break out the logarithmic functions that are there?
 A: HINT 1. $(a-b)(a+b) = a^2-b^2$ for any numbers $a$ and $b$. 
HINT 2. Note that
$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right) = \frac{(\sqrt{3}+\sqrt{2})^2}{3-2}.$$
A: $$\log_b \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = 2\log_b(\sqrt{3}+\sqrt{2}).$$
Since $2\log_b(\sqrt{3}+\sqrt{2}) = \log_b((\sqrt{3}+\sqrt{2})^2)$ and $\log_b$ is a one-to-one function, you should ask whether $(\sqrt{3}+\sqrt{2})/(\sqrt{3}-\sqrt{2})$ is the same as $(\sqrt{3}+\sqrt{2})^2$.  And for that you can rationalize the denominator.
A: $$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$$ 
is equivalent to 
$$\log_b(\sqrt{3}+\sqrt{2}) = \log_b(\sqrt{3}-\sqrt{2}) + 2\log_b(\sqrt{3}+\sqrt{2})$$ 
which is equivalent to 
$$\log_b(\sqrt{3}+\sqrt{2}) = \log_b((\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2)$$ 
which you can multiply out to check, or you might spot $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=\sqrt{3}^2-\sqrt{2}^2=1$.
