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Compute:

$$\frac{1-\log_a^{3}{b} }{(\log_a b+\log_b a+1)\log_a\frac{a}{b}}$$

I tried to expand it :

$$\frac{1-\log_a^{3}{b} }{(\log_a b+\log_b a+1)\log_a\frac{a}{b}}$$

$$=\frac{(1-\log_a{b})(\log_a^{2}b+\log_a b+1)}{(\log_a b+\log_b a+1)(1-\log_a{b})}$$

$$=\frac{(\log_a^{2}b+\log_a b+1)}{(\log_a b+\log_b a+1)}$$

But I got nothing.

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Do you mean $\log b_a$ to be what other people write as $\log_a b$? –  Thomas Andrews May 7 '12 at 13:23
    
Yes sorry , I messed up with the latex. –  Felipe May 7 '12 at 13:24

3 Answers 3

up vote 4 down vote accepted

I think this should help you.

$\log_ba=\frac{\log_aa}{\log_ab}=\frac1{\log_ab}$

Can you finish it from here?

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It worked . @SimonMarkett Yes,I got that. Thank you. –  Felipe May 7 '12 at 13:38

$$ \large{x = \log_a b}$$

Then $$ 1- \log_a^{3}{b} = (1-x^3) $$

Also

$$(\log_a b+\log_b a+1)(\log_a\frac{a}{b}) = (x+\frac{1}{x}+1)(1-x)$$

because

$$ \log_b a = \frac{1}{\log_a b} \hspace{8pt} \textit{and} \hspace{8pt} \log_a \frac{a}{b} = (1-\log_a b)$$

The whole thing gets simplified to

$$ \frac{(1-x^3)}{(x+\frac{1}{x}+1)(1-x)} = \frac{x(1-x^3)}{1-x^3}= x$$

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You also could use $$log_a\frac ab=log_aa-log_ab=1-log_ab$$ Together with @Mike's hint $$log_ba=\frac 1{log_ab}$$ You can express everything in terms of $log_ab$. If you substitute this for readability by, say, $x$ the rest is basic. And to test your result it should be just $log_ab$.

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