Question

log 12 base 3 divided log 3 base 36 - log 4 base 3 divided log 3 base 108​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{\dfrac{log_{3}12}{log_{36}3}-\dfrac{log_{3}4}{log_{108}3}}$

$\underline{\textsf{To find:}}$

$\textsf{The value of the given expression}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{log_{3}12}{log_{36}3}-\dfrac{log_{3}4}{log_{108}3}}$

$\textsf{Using}\;\boxed{\mathsf{log_ab=\dfrac{1}{log_ba}}}$

$\mathsf{=log_{3}12\;log_{3}36-log_{3}4\;log_{3}108}$

$\mathsf{=log_{3}(4{\times}3)\;log_{3}(4{\times}9)-log_{3}4\;log_{3}(4{\times}27)}$

$\textsf{Using product rule of logarithm}$

$\mathsf{=(log_{3}4+log_{3}3)(log_{3}4+log_{3}9)-log_{3}4(log_{3}4+log_{3}27)}$

$\mathsf{=(log_{3}4+log_{3}3)(log_{3}4+log_{3}3^2)-log_{3}4(log_{3}4+log_{3}3^3)}$

$\textsf{Using power rule of logarithm}$

$\mathsf{=(log_{3}4+log_{3}3)(log_{3}4+2\,log_{3}3)-log_{3}4(log_{3}4+3\,log_{3}3)}$

$\mathsf{=(log_{3}4+1)(log_{3}4+2(1))-log_{3}4(log_{3}4+3(1))}$

$\mathsf{=(log_{3}4+1)(log_{3}4+2)-log_{3}4(log_{3}4+3)}$

$\mathsf{=(log_{3}4)^2+3\,log_{3}4+2)-((log_{3}4)^2+3\,log_{3}4)}$

$\mathsf{=(log_{3}4)^2+3\,log_{3}4+2-(log_{3}4)^2-3\,log_{3}4}$

$\mathsf{=2}$

$\implies\boxed{\mathsf{\dfrac{log_{3}12}{log_{36}3}-\dfrac{log_{3}4}{log_{108}3}=2}}$

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Answer 2

Answer:

Your Answer Is 2

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