MHB How to Solve a Logarithmic Equation?

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The discussion focuses on evaluating the expression $\log_{12}18\log_{24}54+5(\log_{12}18-\log_{24}54)$. Participants are encouraged to refer to the Problem of the Week guidelines for submission procedures. The correct solution was provided by user kaliprasad, who received congratulations for their accurate response. The thread emphasizes the importance of understanding logarithmic properties in solving such equations. Engaging with the community can enhance problem-solving skills in mathematics.
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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!
 
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Congratulations to kaliprasad for his correct solution(Cool), which you can find it below:

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