How to solve this exponential equation? I'm given this equation and i have to solve for the x.
$$ e^{2x} -(e^5 + e^2)e^x + e^7 = 0      $$
The results should be $x =2$ and $x = 5$. 
At first i thought it would be an easy task, substituting $e^x$ with $y$, but I can't get rid of the coefficients. I'm sorry if this question is too simple for this site but I'd be really grateful if you could explain me. Thanks for your kindness.
 A: Any quadratic can be written as $$x^2-(\text{sum of roots})x+(\text{product of roots})=0$$
A: You can make $e^{2x}-(e^2+e^5)e^x+e^7=(e^x-e^2)(e^x-e^5)=0$. So we have the result.
A: You are on the right path! Making the substitution you mentioned, will lead us to the quadratic equation:
$$y^2  - (e^5+e^2)\cdot  y + e^7 = 0.\tag {1}$$
After finding $y_1, y_2$, we have to go back, set $y_i =  e^x, \, (i = 1,2)$ and solve for $x$.

Recall that if we have the quadratic equation 
$$ay^2 + by + c = 0,$$ then the discriminant is given by the formula 
$D = b^2 - 4ac$ and I hope you remember the formula that gives the solution(s). Can you guess what the coefficients $a,b,c$ in equation $(1)$ are?
A: When given quadratic expression is factorized,  $(e^x-e^2) \cdot (e^x-e^5)=0$ 
Just equate the exponents, $ x= 2, 5. $
A: We have $$e^{2x}-(e^5+e^2)e^x+e^7=0$$ $$(e^{x})^2-(e^5+e^2)e^x+e^7=0$$ Now, solving above quadratic equation for $e^{x}$ $$e^x=\frac{-(-(e^5+e^2))\pm\sqrt{(e^5+e^2)^2-4(1)(e^7)}}{2\cdot 1}$$ $$=\frac{(e^5+e^2)\pm\sqrt{(e^5+e^2)^2-4(1)(e^7)}}{2\cdot 1}$$ $$=\frac{e^2(e^3+1)\pm e^2\sqrt{e^6+1-2e^3}}{2}=\frac{e^2(e^3+1)\pm e^2\sqrt{(e^3-1)^2}}{2}$$ $$=\frac{e^2(e^3+1)\pm e^2(e^3-1)}{2}$$ $\color{red}{\text{Taking positive sign}}$ $$e^x=\frac{e^2(e^3+1)+ e^2(e^3-1)}{2}=\frac{2e^5}{2}=e^5\iff x=5$$ $\color{red}{\text{Taking negative sign}}$ $$e^x=\frac{e^2(e^3+1)- e^2(e^3-1)}{2}=\frac{2e^2}{2}=e^2\iff x=2$$ Hence, we have two values of $x$ given as $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=2\ \text {&}\ x=5}}$$
A: We have,
$$
e^{2x}-(e^5+e^2)e^x+e^7=0
$$
Note that it 'looks' a bit like a quadratic equation. Since these are easy to solve, we should try and put it into a simpler form.
$$
(e^x)^2-(e^5+e^2)e^x+e^7=0
$$
Let's make the substitution: $$u=e^x$$
This gives us:
$$
u^2 - (e^5+e^2)u + e^7 = 0
$$
This can then be factorised to:
$$
(u-e^5)(u-e^2)=0
$$
This gives us two solutions:
$$
u-e^5 = 0 \\
e^x=e^5 \\
x = 5
$$
or
$$
u-e^2 = 0 \\
e^x=e^2 \\
x = 2
$$
