Hi Guest!
\(f(x)=x2^x \\ \implies f'(x)=\dfrac{d}{dx}[x2^x]=\dfrac{d}{dx}(x)*2^x+x\dfrac{d}{dx}2^x \hspace{0.5cm} \text{(Product rule)} \\ \implies f'(x)=2^x+x2^x ln(2)\)
Next, we set: \(f(x)=2f'(x)\) to get:
\(x2^x=2[x2^xln(2)+2^x] \\ \iff x=2ln(2)x+2 \\ \iff x-ln(4)x=2 \\ x=\dfrac{2}{1-ln(4)}\)
which is the only real solution of x.
I hope this helps.
