How to Solve a Logarithmic Equation?

[anemone]

Here is this week's POTW:

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Evaluate $\log_{12}18\log_{24}54+5(\log_{12}18-\log_{24}54)$.-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution(Cool), which you can find it below:

Spoiler

Let us take $x= \log\, 2$ and $y=\log\,3$

Let $A =\dfrac{\log\,18}{\log\,12} = \dfrac{\log\,2 + 2\log\,3}{2\log\,2 + \log\,3} = \dfrac{x+2y}{2x+y}$

and $B =\dfrac{\log\,54}{\log\,24} = \dfrac{\log\,2 + 3\log\,3}{3\log\,2 + \log\,3} = \dfrac{x+3y}{3x+y}$

Now

AB + 5(A - B)

$=(A-5)(B+5) + 25$

$=\left(\dfrac{x+2y}{2x+y}-5\right)\left(\dfrac{x+3y}{3x+y} + 5\right)+25$

$=\left(\dfrac{-9x-3y}{2x+y}\right)\left(\dfrac{16x+8y}{3x+y}\right)+25$

$=\left(\dfrac{-3(3x+y)}{2x+y}\right)\left(\dfrac{8(2x+y)}{3x+y}\right)+25$

$= - 24 + 25 \\=1$

Therefore, $\log_{12}18\log_{24}54+5(\log_{12}18-\log_{24}54)=1$.