What are the two solutions to this equation? $$e^{2x}-21e^x+110=0$$ I'm stumped! Taking $\log$ of both sides doesn't seem to work?

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    $\begingroup$ Set $y=e^x$ and solve as a regular quadratic (since $e^{2x}=y^2$). Once you have $y$, take $x=ln(y)$. $\endgroup$ – user2825632 Sep 17 '16 at 4:27
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Sep 17 '16 at 4:30
  • $\begingroup$ How about a u-substitution with $u=e^{x} $. $\endgroup$ – Joe Sep 17 '16 at 4:54

Hint: Write the equation as $$(e^x)^2-21(e^x)+110=0$$ and solve for $e^x$. Then solve the resulting equations for $x$.

You can simply factor as $$(e^x-10)(e^x-11)=0$$ and go from there.

So now you should be able to solve $ae^{2x}+be^x+c=0$.



$y=e^{x}$, so $y^{2}=e^{2x}$, and we had $e^{2x}-21e^{x}+110=0$

So: $$y^{2}-21y+110=0$$

This is Quadratic Equation and if we solve this (for example with Quadratic formula), two values will be calculated for $y$:

$$y_{1}=10\\ y_{2}=11$$

At first we set $y=e^{x}$, so $x=ln(y),\ y>0$. $y_1$ and $y_2$ are positive ($y_1, y_1>0$) so there are two values for $x $, $x_1 $ and $ x_2$:

$$x_1 = \ln(y_1) = \ln(10)$$ $$x_2 = \ln(y_2) = \ln(11)$$


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