Basic of Logarithms
easy

If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=

A

$0$

B

$1$

C

$x + y + z$

D

$x + y + z + 2$

Solution

(d) ${a^x}.{b^y}.{c^z} = bc.ca.ab = {a^2}{b^2}{c^2}$

$ \Rightarrow $${a^{x – 2}}{b^{y – 2}}{c^{z – 2}} = 1 = {a^0}{b^0}{c^0}$

$\therefore x = y = z = 2$

$\therefore xyz = {2^3} = 8 = x + y + z + 2$.

Standard 11
Mathematics

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