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(xlogx⋅x2)=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(gof)(x)=g(f(x))=g(xlogx⋅x2)=g(x2+logx)=log(x2+logx)+1
Now, (gof)(x)=1(gof)(x)=1(gof)(x)=1
⇒log(x2+logx)+1=1⇒log(x2+logx)=0⇒x2+logx=1⇒x2+logx=x0⇒2+logx=0⇒logx=−2⇒x=e−2\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}⇒log(x2+logx)+1=1⇒log(x2+logx)=0⇒x2+logx=1⇒x2+logx=x0⇒2+logx=0⇒logx=−2⇒x=e−2
Hence, real solution is x=e−2x=e^{-2}x=e−2
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