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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