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!