(e^x)-(15e^-x)=2 solve for x

Question

I mainly just can't figure out the -x in 15e^-x. (This is actually precalculus work.) So if you would kindly show me how to do it! Thank you!

Answer 1

This can be written: e^2x-2e^x-15=0; (e^x-5)(e^x+3)=0, so e^x=5 or -3 and x=ln(5) or ln(-3). Since we can't have the log of a negative number, the only solution is x=ln(5)=1.6094 approx.

Answer 2

e^x - 15e^-x - 2 = 0

Let's give e^x a value of y:

y - 15/y - 2 = 0

Multiply everything by y to cancel off the denominator:

y^2 - 2y - 15 = 0

Factor it out:

(y - 5)(y + 3) = 0

Solve for each:

y - 5 = 0
y = 5

y + 3 = 0
y = -3

Now substitute e^x back for y and solve for each:

e^x = 5
x = ln5

Get detailed explanation on Precalculus Help