frac{log _{9} 11}{log _{5} 13}-frac{log _{3} | vedclass

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(C) We know that $\log_{a^n} b = \frac{1}{n} \log_a b$ and $\log_{a^{1/n}} b = n \log_a b$.First,simplify the terms:$\log_9 11 = \log_{3^2} 11 = \frac

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(C) We know that $\log_{a^n} b = \frac{1}{n} \log_a b$ and $\log_{a^{1/n}} b = n \log_a b$.First,simplify the terms:$\log_9 11 = \log_{3^2} 11 = \frac{1}{2} \log_3 11$.$\log_{\sqrt{5}} 13 = \log_{5^{1/2}} 13 = 2 \log_5 13$.Substituting these into the expression:$\frac{\frac{1}{2} \log_3 11}{\log_5 13} - \frac{\log_3 11}{2 \log_5 13} = \frac{1}{2} \frac{\log_3 11}{\log_5 13} - \frac{1}{2} \frac{\log_3 11}{\log_5 13} = 0$.

$\frac{\log _{9} 11}{\log _{5} 13}-\frac{\log _{3} 11}{\log _{\sqrt{5}} 13}=$

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