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} \u00b7 x^2)\n\\\\=g(x^{2+\\log x})\n\\\\=\\log (x^{2+\\log x})+1"
Now, "(gof)(x)=1"
"\\Rightarrow \\log (x^{2+\\log x})+1=1\n\\\\\\Rightarrow \\log (x^{2+\\log x})=0\n\\\\\\Rightarrow x^{2+\\log x}=1\n\\\\\\Rightarrow x^{2+\\log x}=x^0\n\\\\ \\\\\\Rightarrow {2+\\log x}=0\n\\\\ \\\\\\Rightarrow \\log x=-2\n\\\\\\Rightarrow x=e^{-2}"
Hence, real solution is "x=e^{-2}"
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