# Exact Solution for Logarithmic Equation?

I am faced with this equation, and I don't really know where to begin:

$$x^2e^2 - 2e^x = 0.$$

I usually start these types of problems by factoring out a common term, but I don't see any in this particular example.

• Are you familiar with mathworld.wolfram.com/LambertW-Function.html? – Amzoti Nov 18 '13 at 3:49
• Amzoti is correct. The Lambert W function the only way to solve this problem. If this came up in an elementary math course, the book or the professor made a typo. – Stephen Montgomery-Smith Nov 18 '13 at 3:51
• Jenny, welcome to math.SE. Luis edited your question and your formula is now rendered in a much more reader friendly way. If you try to edit you question, you will see that your former formula is no encapsulated between dollar signs, which allow to enter Mathjax code and produce this nice rendering. For an introduction to mathjax code, you may want to look at meta.math.stackexchange.com/questions/5020/… – Martin Van der Linden Nov 18 '13 at 3:55

$$e^2x^2-2e^x =0 \implies \left( x-\frac{\sqrt{2}}{e}e^{x/2} \right)\left( x+\frac{\sqrt{2}}{e}e^{x/2} \right)=0.$$
$$\implies \left( x-\frac{\sqrt{2}}{e}e^{x/2} \right)=0 \quad \rm{or}\quad \left( x+\frac{\sqrt{2}}{e}e^{x/2} \right)=0.$$
i) make the change of variables $\frac{x}{2}=u$