I am a beginner to logarithms, and I need to know how to simplify the following expression: $$ \dfrac{(a^3-1)(b^3-1)-1}{ba(a+b)} \;, $$ given that $\log(a+b) = \log(ab)$ for $a,b > 0$.
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2The logarithm is a 1-1 function over its domain, so $log(a+b) = log(ab)$ implies $a+b = ab$. Can you use this to start the simplification? – Osama Ghani Jul 02 '17 at 07:24
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@Osama Ghani I'm starting – Fghj Jul 02 '17 at 07:28
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@Osama Ghani What do you mean by saying 1-1 function? What is the domain of logarithmic functions? – Fghj Jul 02 '17 at 08:04
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It just means that if $log(a+b) = log(ab)$ then $(a+b) = ab$. – Osama Ghani Jul 02 '17 at 08:06
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That I know, but I am willing to know about that things which I've asked above – Fghj Jul 02 '17 at 08:10
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1A 1-1 function is one such that if $f(x) = f(y)$ then $x = y$. The domain just means what the function can take as an input. So here the domain of $log(x)$ is $x > 0$. – Osama Ghani Jul 02 '17 at 08:13
2 Answers
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Let the answer be some t.then apply log on both sides.we know that log (a/b)=loga-love and log (ab)=loga+logb using both these formulae continue solving.later sustitute the formula for a^3+b^3.
mechanics
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When $a,b>0,\ \ln(a+b)=\ln(ab)\iff a+b=ab$
$\displaystyle \frac{(a^3-1)(b^3-1)-1}{ab(a+b)}=\frac{a^3b^3-a^3-b^3}{ab(a+b)}=\frac{(a+b)^3-a^3-b^3}{ab(a+b)}=\frac{3a^2b+3ab^2}{ab(a+b)}=\frac{3ab(a+b)}{ab(a+b)}=3$
zwim
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