Logarithm integrated with Quadratics

[Leah5467]

Hello,this is the question i don't know:
6d

Thank you!

 

[HallsofIvy]

You yourself use the word "quadratics" in your title! First do you know that $\displaystyle e^{2x}= (e^x)^2$?
So (a) $\displaystyle e^{2x}= 2e^x$ is $\displaystyle (e^x)^2= 2e^x$. If you like you can let $\displaystyle y= e^x$ so that your equation becomes $\displaystyle y^2= 2y$. Solve that for y then solve $\displaystyle e^x= y$ for x.

For (d) first multiply both sides by $\displaystyle e^x$: $\displaystyle (e^x)^2+ 2e^x= 3$. Again let $\displaystyle y= e^x$ so that the equation becomes $\displaystyle y^2+ 2y= 3$.

 

[Leah5467]

Thank you!