Given-log-2-10-a-a-1-log-3-5-1-b-find-the-value-of-1-log-12-15-

Question Number 120380 by bramlexs22 last updated on 31/Oct/20

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

{\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}