How do you solve ex15ex2=0?

1 Answer
Oct 14, 2015

x=ln(5)

Explanation:

Multiply both sides by ex

ex+x15exx2ex=0
e2x152ex=0

Call ex=y so we have

y2152y=0
y22y15=0

Solve the quadratic

y=2±441(15)2

y=2±4+602

y=2±82

y=1±4

y=5 or y=3

But y=ex so we really have

ex=5 or ex=3

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

ex=5

Taking the log of both sides we have

x=ln(5)