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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?

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    Good suggestion, starting over with just the left hand side.2011-11-29

3 Answers 3

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HINT 1. $(a-b)(a+b) = a^2-b^2$ for any numbers $a$ and $b$.

HINT 2. Note that $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right) = \frac{(\sqrt{3}+\sqrt{2})^2}{3-2}.$

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    It was not copied correctly, the right hand side has been edited above to say: $2\log_b(\sqrt{3}+\sqrt{2})$2011-11-29
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$\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.

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$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$ 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 $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=\sqrt{3}^2-\sqrt{2}^2=1$.