Question

log root a to the base b into log cube root b to the base e into log fourth root of c to the base is equal to

Answer 1

$\textbf{Given:}$

$\mathsf{log_b\sqrt{a}{\times}log_c\sqrt[3]{b}{\times}log_a\sqrt[4]{c}}$

$\textbf{To find:}$

$\textsf{The value of}$

$\mathsf{log_b\sqrt{a}{\times}log_c\sqrt[3]{b}{\times}log_a\sqrt[4]{c}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{log_b\sqrt{a}{\times}log_c\sqrt[3]{b}{\times}log_a\sqrt[4]{c}}$

$\mathsf{=log_ba^\frac{1}{2}{\times}log_cb^\frac{1}{3}{\times}log_ac^\frac{1}{4}}$

$\mathsf{Using,}$

$\boxed{\mathsf{log_aM^n=n\;log_aM}}$

$\mathsf{=\dfrac{1}{2}\;log_ba{\times}\dfrac{1}{3}log_cb{\times}\dfrac{1}{4}log_ac}$

$\mathsf{=\dfrac{1}{24}\;(log_ba{\times}log_cb{\times}log_ac)}$

$\mathsf{Using,}$

$\boxed{\mathsf{log_ab{\times}log_bc=log_ac}}$

$\mathsf{=\dfrac{1}{24}\;(log_bc{\times}log_cb)}$

$\mathsf{=\dfrac{1}{24}\;(log_bb)}$

$\mathsf{=\dfrac{1}{24}\;(1)}$

$\mathsf{=\dfrac{1}{24}}$

$\implies\boxed{\mathsf{log_b\sqrt{a}{\times}log_c\sqrt[3]{b}{\times}log_a\sqrt[4]{c}=\dfrac{1}{24}}}$

$\textbf{Find more:}$

Simply log10( 145/8)-log10(3/2)+log10(54/29)​

https://brainly.in/question/33031758

Answer 2

Answer:

log root a to the base b into log cube root b to the base e into log fourth root of c to the base is equal to

Step-by-step explanation:

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