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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