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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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Suggestion: Don't manipulate the entire "equality" unless you already know it is true; if you start with a false equality you can still end up with something "true", if your steps are not reversible. E.g., starting from $1=2$ you can easily end up with $1=1$. –  Arturo Magidin Nov 29 '11 at 22:46
    
Good suggestion, starting over with just the left hand side. –  erimar77 Nov 29 '11 at 22:50

3 Answers 3

up vote 5 down vote accepted

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: $$2\log_b(\sqrt{3}+\sqrt{2})$$ –  erimar77 Nov 29 '11 at 22:52

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

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

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