How can I find y?

Question

The following equation is given: $$ye^y=e^{x+1}$$ when $x=0$. I tried to solve it as logarithmic equation but I can't go further. I know $y=1$ but I don't know how to prove it. Any idea? Thank you

Answer 1

You may write

• $ye^y = e^{\ln y + y} = e^{x+1} \Rightarrow \ln y + y = 1+x \stackrel{x= 0}{\Rightarrow} \Rightarrow \ln y + y = 1 \Rightarrow y =1$

The uniqueness of the solution follows from the monotonicity of $\ln y+ y$.

Answer 2

If $x=0$, then $y=1$ is obviously a solution. But $e^{x+1}=e\cdot e^x>e^x$, and $x+1>x$, hence $xe^x$ is an increasing function. Since $y=1$ solves $ye^y=e$, and $e$ is constant, it is unique.

Answer 3

You can solve this equation by taking logarithms both side with respect to the base $e$. The equation becomes: $$\ln(ye^y)=\ln(e^{x+1})$$ $$\ln y+\ln e^y=\ln (e^{x+1})$$ $$\ln y +y = x+1$$ You are given the value of $x=0$, put it in the equation to get: $$\ln y +y =1$$

From the equation, if we put $y=1$ it satisfies the equation. You can also plot a graph for the two functions.

Hope this helps...