How do you solve #e^x - 15e^(-x) - 2 = 0 #?

Question

6675 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-14T23:04:32

#x = ln(5)#

Multiply both sides by #e^x#

#e^(x+x) - 15e^(x-x) - 2e^x = 0#
#e^(2x) - 15 - 2e^x = 0#

Call #e^x = y# so we have

#y^2 - 15 - 2y = 0#
#y^2 - 2y - 15 = 0#

Solve the quadratic

#y = (2 +-sqrt(4 - 4*1*(-15)))/2#

#y = (2 +-sqrt(4 +60))/2#

#y = (2 +- 8)/2#

#y = 1 +- 4#

#y = 5# or #y = -3#

But #y = e^x# so we really have

#e^x = 5# or #e^x = -3#

The exponential function is positive for every real #x# so we can ignore that root and say that

#e^x = 5#

Taking the log of both sides we have

#x = ln(5)#