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Prove that if $\alpha = \log_{12}18$ and $ \beta = \log_{24}54$ then $ \alpha \beta +5(\alpha - \beta)=1$

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Do you know log properties, e.g., $\log_k(b)\log_b(x)=\log_k(x)$? –  snarski Feb 18 '13 at 2:57
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Is that an order? –  copper.hat Feb 18 '13 at 2:57
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@copper I assume you're asking the OP? If so, absolutely. chop chop. –  snarski Feb 18 '13 at 2:58
    
@snarski: Indeed, the comment was aimed at the OP! –  copper.hat Feb 18 '13 at 2:59
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Sachin, where did you get this problem? –  Will Jagy Feb 18 '13 at 20:05

2 Answers 2

Suppose we have positive integers $m,n$ with $m^2 + mn - n^2 = 1.$ This means they are consecutive fibonacci numbers $m = f_{2k}, \; \; n = f_{2k+1},$ for example $(m=1,n=1), \; \; (m=2,n=3), \; \; (m=5,n=8), \; \; (m=13,n=21). $ Next suppose we have real or complex variables $x,y.$ Next we take $$ \alpha = \frac{x + m y}{m x + y}, \; \; \; \beta = \frac{x + n y}{n x + y}. $$ Then, by putting on a common denominator, we can confirm that $$ \alpha \beta + (m+n)\alpha - (m+n) \beta = 1. $$ This is really very clever and not something I knew. I am wondering what other such things might be true, combining indefinite quadratic forms with rational functions in two variables that resemble linear fractional transformations.

As @lab has pointed out, the values of $x,y$ do not matter at all. However, for this problem they can be taken to be natural logarithms $x = \log 2, y = \log 3.$ Oh, for this problem $m=2,n=3.$

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$$\log_{12}18=\frac{\log_218}{\log_212}=\frac{\log_22+\log_23^2}{\log_22^2+\log_23}=\frac{1+2\log_23}{2+\log_23}$$ as $\log_ab=\frac{\log_cb}{\log_ca}$ where $a\ne1,b,c\ne1$ are positive real numers and $\log mn=\log m+\log n$

Similarly, simplify $\log_{24}54$ and equate the values of $\log_23$

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