Basic of Logarithms
medium

If ${a^x} = b,{b^y} = c,{c^z} = a,$ then value of $xyz$ is

A

$0$

B

$1$

C

$2$

D

$3$

Solution

(b) ${a^x} = b \Rightarrow $ $x\log a = \log b$

==> $x = {{\log b} \over {\log a}} = {\log _a}b$

Similarly $y = {\log _b}c,\,z = {\log _c}a$

$\therefore xyz = {\log _a}b.{\log _b}c.{\log _c}a = 1$.

Standard 11
Mathematics

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