Given { ((log _2 (10)=(a/(a−1)))),((log _3 (5)=(1/b))) :} find the value of 1+log


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Question Number 120380 by bramlexs22 last updated on 31/Oct/20
$Given { ((log _2 (10)=(a/(a−1)))),((log _3 (5)=(1/b))) :} find the value of 1+log _(12) (15) ?$
$G i v e n {\begin{cases} \log_{2} (10) = \frac{a}{a - 1} \\ \log_{3} (5) = \frac{1}{b} \end{cases}$ $f i n d t h e v a l u e o f 1 + \log_{12} (15) ?$
	Answered by FelipeLz last updated on 31/Oct/20

	$\log_{2} 10 = 1 + \frac{\ln 5}{\ln 2} = \frac{a}{a - 1} \Rightarrow \ln 2 = (a - 1) \ln 5$ $\log_{3} 5 = \frac{\ln 5}{\ln 3} = \frac{1}{b} \Rightarrow \ln 3 = b \ln 5$ $1 + \log_{12} 15 = 1 + \frac{\ln 3 + \ln 5}{\ln 3 + 2 \ln 2} = 1 + \frac{b \ln 5 + \ln 5}{b \ln 5 + 2 (a - 1) \ln 5} = 1 + \frac{b + 1}{b + 2 (a - 1)} = \frac{2 (a + b) - 1}{2 (a - 1) + b}$
	Answered by john santu last updated on 31/Oct/20
	${ ((log _2 (5)+1 = (a/(a−1)) ⇒log _2 (5)=((a−a+1)/(a−1)) = (1/(a−1)))),((log _3 (5)=(1/b))) :} ⇒1+((log _5 (3)+1)/(log _5 (3)+2log _5 (2))) = 1+((b+1)/(b+2a−2)) ⇒ ((2a+2b−1)/(2a+b−2))$
	${\begin{cases} \log_{2} (5) + 1 = \frac{a}{a - 1} \Rightarrow \log_{2} (5) = \frac{a - a + 1}{a - 1} = \frac{1}{a - 1} \\ \log_{3} (5) = \frac{1}{b} \end{cases}$ $\Rightarrow 1 + \frac{\log_{5} (3) + 1}{\log_{5} (3) + 2 \log_{5} (2)} = 1 + \frac{b + 1}{b + 2 a - 2}$ $\Rightarrow \frac{2 a + 2 b - 1}{2 a + b - 2}$
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