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Basic of Logarithms
medium
If ${x^y} = {y^x},$then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $
A
$0$
B
$1$
C
$-1$
D
None of these
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
