Going around in circles with this.

[Joffa]

I’ve got the answer but I can’t get it myself. I’ve been going around in circles for hours. I have done all the other questions on this sheet without too many problems but i’m Stuck on this one and there’s some thing i’m Just not seeing. Q: 4/(2+e^x) = 6/(3+e-x). Solve for x.
I’ve tried multiplying both sides to get rid of the bases. I’ve tried inverting the original fractions. I’ve tried 1/4(2+e^x)^-1=1/2(3+e^-x)^-1 but I always seem to get muddled up somewhere and can’t get to the solution. Please help!

 

[Prof]

It sounds like you have done quite a bit of work on this problem. You should submit what you think is your best effort, and we can look for where you made an error. You could start by multiplying both sides by the common denominator (setting the cross products equal which is the same thing). You said "I tried multiplying both sides to get rid of the bases," so tell us step-by-step what happened when you did that. If you are able to, you could photograph your paper.

 

[Dr.Peterson]

After multiplying by both denominators to clear fractions, you should have found an equation with a term containing e^x, one containing e^-x, and another. The step you are likely to have missed is to multiply that equation by e^x. Did you do that? Did you see why it helps?

(Actually, it's a bit easier than what I just described; I said that without having done it.)

As Prof said, we need to see your work in order to know what will help you.