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Basic of Logarithms
medium
If ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ then
A
$x = y = z = a/3$
B
$x + y + z = a/3$
C
$x + y + z = 0$
D
None of these
Solution
(a) ${a^x}.{a^y}.{a^z} = {(x + y + z)^{y + z + x}}$
$ \Rightarrow \,\,{a^{x + y + z}} = {(x + y + z)^{x + y + z}}$$ \Rightarrow $ $x + y + z = a$
Now, ${a^x} = {(x + y + z)^y} = {a^y}$$ \Rightarrow $ $x = y$, similarly $y = z$
$\therefore x = y = z = {a \over 3}$.
Standard 11
Mathematics
