Answer to Question #247206 in Discrete Mathematics for Alina

Question #247206

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



1
Expert's answer
2021-10-12T10:31:28-0400

Solution:

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

(gof)(x)=g(f(x))=g(xlogxx2)=g(x2+logx)=log(x2+logx)+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(gof)(x)=1

log(x2+logx)+1=1log(x2+logx)=0x2+logx=1x2+logx=x02+logx=0logx=2x=e2\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=e2x=e^{-2}


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