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$\begingroup$

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.

  • 0
    Yes sorry , I messed up with the late$x$.2012-05-07

3 Answers 3

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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.2012-05-07
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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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$ \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$