If log245175=a, log1715875=b, then the value of 1−aba−b is ____.

a=log245175=log175log245=log(52×7)log(5×72)=2log5+log7log5+2log7=2α+βα+2β a=log1715875=log875log1715=log(53×7)log(5×73)=3log5+log7log5+3log7=3α+βα+3β Where α=log5 and β=log7 Substituting a and b Now, 1−aba−b=5(β2−α2)β2−α2=5

If log245175=a, log1715875=b, then the value of 1−aba−b is ____.

a=log245175=log175log245=log(52×7)log(5×72)=2log5+log7log5+2log7=2α+βα+2β a=log1715875=log875log1715=log(53×7)log(5×73)=3log5+log7log5+3log7=3α+βα+3β Where α=log5 and β=log7 Substituting a and b Now, 1−aba−b=5(β2−α2)β2−α2=5

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If $\log_{245} 175 = a, \log_{1715} 875 = b,$ then the value of $\frac{1 - a b}{a - b}$ is ____.

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answer is D.

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Detailed Solution

$a = \log_{245} 175 = \frac{\log 175}{\log 245} = \frac{\log (5^{2} \times 7)}{\log (5 \times 7^{2})} = \frac{2 \log 5 + \log 7}{\log 5 + 2 \log 7} = \frac{2 α + β}{α + 2 β}$
$a = \log_{1715} 875 = \frac{\log 875}{\log 1715} = \frac{\log (5^{3} \times 7)}{\log (5 \times 7^{3})} = \frac{3 \log 5 + \log 7}{\log 5 + 3 \log 7} = \frac{3 α + β}{α + 3 β}$
Where $α = \log 5$ and $β = \log 7$
Substituting a and b
Now, $\frac{1 - a b}{a - b} = \frac{5 (β^{2} - α^{2})}{β^{2} - α^{2}} = 5$