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.
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.
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?
Any quadratic can be written as $$x^2-(\text{sum of roots})x+(\text{product of roots})=0$$
When given quadratic expression is factorized, $(e^x-e^2) \cdot (e^x-e^5)=0$
Just equate the exponents, $ x= 2, 5. $
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 $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=2\ \text {&}\ x=5}}$$
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
$$ u-e^2 = 0 \\ e^x=e^2 \\ x = 2 $$
