Solve for x if (g ◦ f)(x) = 1. Here, f(x) = (xlog(x) · x2) and g(x) = log(x) + 1.

Solution:

 f(x) = (xlog(x) · x2) and g(x) = log(x) + 1.

$(gof)(x)=g(f(x))=g(x^{\log x} · x^2) \\=g(x^{2+\log x}) \\=\log (x^{2+\log x})+1$

Now, $(gof)(x)=1$

$\Rightarrow \log (x^{2+\log x})+1=1 \\\Rightarrow \log (x^{2+\log x})=0 \\\Rightarrow x^{2+\log x}=1 \\\Rightarrow x^{2+\log x}=x^0 \\ \\\Rightarrow {2+\log x}=0 \\ \\\Rightarrow \log x=-2 \\\Rightarrow x=e^{-2}$

Hence, real solution is $x=e^{-2}$