Expert: Sherry Wallin - 7/23/2009

Question
QUESTION: solve for x: e^(2x) - 2e^(x+2) + 4e = 0 (hint: factor)

ANSWER: Sabrina u substitution works nicely for this one. Let u = e^x so your equation becomes u^2-(2e^2)u 4e = 0.
Factor by using the quadratic formula or complete the square and you will get that u = e^2  -sqrt(e^4-4e).
Notice this is just a number because e is just a number. Now replace u with e^x so you have
e^x =e^2  -sqrt(e^4-4e)

Take the ln of both sides and get that x = ln(e^2  -sqrt(e^4-4e)).

When you crunch those values of x into a calculator to check your values of x you find that for the x = ln(e^2 + sqrt(e^4-4e)) you get -.03, which is close to 0.  For x = ln(e^2 -sqrt(e^4-4e)) you get back about .0013 (which is also very close to 0). This is only an approximation because e is irrational so this is as good as it gets!

Math Prof

---------- FOLLOW-UP ----------

QUESTION: made a mistake with the original problem it is e^(2x)-2e^(x+2)+e^4 = 0 (hint:factor)

sorry about the type-o

Answer
This makes the problem even easier yet since it is not necessary to use the quadratic formula or to complete the square to factor it. The first thing I would do is divide through by e^4:
(e^2x -2e^(x+2)+e^4)/e^4 = e^(2x-4)-2e(x-2)+ 1 = e^[2(x-2)]-2e(x-2)+ 1. Now use u substitution letting u = e^(x-2) getting (u^2-2u+1) which factors easily as (u-1)^2, so set (u-1)=0 -> u = 1 and plug back in for u = e^(x-2) so you have e^(x-2) = 1. Take the natural log of both sides getting:
ln[e^(x-2)] = ln 1 -> x-2 = 0 or x = 2. This is a double root so check it in the original equation to make sure it makes 0: e^(2*2) -2e^4 + e^4 = e^4 - e^4 = 0. It works so x = 2 is the solution.

Math Prof