Logarithm

Posted June 15th, 2009 by alpha

If $log_{12} 18=\alpha$ [log 18 to the base 12] and $log_{24} 54=\beta$ [log 54 to the base 24] , Prove that $\alpha\beta+5(\alpha-\beta)=1$

Please show your steps.

$log_2{3}=t$

$a=\frac{2t+1}{t+2}$

$b=\frac{3t+1}{t+3}$

$\frac{2t+1}{t+2}\frac{3t+1}{t+3}+5(\frac{2t+1}{t+2}-\frac{3t+1}{t+3})=1$

Comments

On June 15th, 2009 Structure says:

$log_2{3}=t$

$a=\frac{2t+1}{t+2}$

$b=\frac{3t+1}{t+3}$

$\frac{2t+1}{t+2}\frac{3t+1}{t+3}+5(\frac{2t+1}{t+2}-\frac{3t+1}{t+3})=1$

On June 15th, 2009 alpha says:

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