Basic of Logarithms
medium

જો ${x^y} = {y^x},$ તો ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ કે જ્યાં $k = . . . . $

A

$0$

B

$1$

C

$-1$

D

એકપણ નહીં

Solution

(b) ${x^y} = {y^x} \Rightarrow {({x^y})^{1/x}} = y$

Now, ${\left( {{x \over y}} \right)^{x/y}} = {\left( {{x \over {{x^{y/x}}}}} \right)^{x/y}} = {\left( {{x^{1\, – \,{y \over x}}}} \right)^{x/y}}$

$={x^{(x/y) – 1}} = {x^{(x/y) – k}} \Rightarrow k = 1$.

Standard 11
Mathematics

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