Question:
If $a=1, b=2$ then find the value of $\left(a^{b}+b^{a}\right)^{-1}$.
Solution:
For a = 1and b = 2,
$\left(a^{b}+b^{a}\right)^{-1}$
$=\left(1^{2}+2^{1}\right)^{-1}$
$=(1+2)^{-1}$
$=3^{-1}$
$=\frac{1}{3} \quad\left(x^{-a}=\frac{1}{x^{a}}\right)$
Thus, the value of $\left(a^{b}+b^{a}\right)^{-1}$ when $a=1$ and $b=2$ is $\frac{1}{3}$.
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