Finding the value of $x$ when ${x\ln(x)=1}$ and ${x>0}$

Question

I have to find the solution to the equation above and show that it's the only solution with a value of somewhere between 1 and 2.

This is what I tried to do:

$$x\ln(x)=1$$ $$\ln(x)^x=1$$ $$e^1=x^x$$ $$e=x^x$$ and then I got lost. Any suggestions?