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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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:$\frac{\frac{1}{2} \log_3 11}{\log_5 13} - \frac{\log_3 11}{2 \log_5 13} = \frac{1}{2} \frac{\log_3 11}{\log_5 13} - \frac{1}{2} \frac{\log_3 11}{\log_5 13} = 0$.

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

Similar Questions

For $x>0,$ if $y=\frac{10^{\log _{10} x}}{x^{2}}$ and $x=y^{a},$ then $a=$

If $\log _{10} 50+\log _{10} 20=x,$ then $x$ is equal to

If $\frac{\log x}{l+m-2n} = \frac{\log y}{m+n-2l} = \frac{\log z}{n+l-2m}$,find the value of $x^2 y^2 z^2$.

If $a^{x} = b^{y}$,then

$\log _{5}\left(1+\frac{1}{5}\right)+\log _{5}\left(1+\frac{1}{6}\right)+\log _{5}\left(1+\frac{1}{7}\right)+\cdots+\log _{5} \left(1+\frac{1}{624}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module