If {log _4}5 = a and {log _5}6 = b, then find | vedclass

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(C) Given: ${\log _4}5 = a$ and ${\log _5}6 = b$. Using the change of base formula,${\log _4}5 = \frac{{\log 5}}{{\log 4}} = \frac{{\log 5}}{{2\log 2}

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$\frac{1}{{2ab - 1}}$

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(C) Given: ${\log _4}5 = a$ and ${\log _5}6 = b$. Using the change of base formula,${\log _4}5 = \frac{{\log 5}}{{\log 4}} = \frac{{\log 5}}{{2\log 2}} = a \implies \frac{{\log 5}}{{\log 2}} = 2a$. Also,${\log _5}6 = \frac{{\log 6}}{{\log 5}} = b \implies \frac{{\log 2 + \log 3}}{{\log 5}} = b$. From the first equation,$\frac{{\log 2}}{{\log 5}} = \frac{1}{{2a}}$. Substitute this into the second equation: $\frac{{\log 2}}{{\log 5}} + \frac{{\log 3}}{{\log 5}} = b \implies \frac{1}{{2a}} + \frac{{\log 3}}{{\log 5}} = b$. $\frac{{\log 3}}{{\log 5}} = b - \frac{1}{{2a}} = \frac{{2ab - 1}}{{2a}}$. Therefore,$\frac{{\log 5}}{{\log 3}} = \frac{{2a}}{{2ab - 1}}$. We need ${\log _3}2 = \frac{{\log 2}}{{\log 3}} = \frac{{\log 2}}{{\log 5}} 	imes \frac{{\log 5}}{{\log 3}} = \frac{1}{{2a}} 	imes \frac{{2a}}{{2ab - 1}} = \frac{1}{{2ab - 1}}$.

If ${\log _4}5 = a$ and ${\log _5}6 = b,$ then find the value of ${\log _3}2$.

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