SOLUTION: Given the equations, rewrite each as a quadratic equations in u, where u = e^x. Solve for u first then for x 1) e^2x - e^x =6 2)e^-2x

SOLUTION: Given the equations, rewrite each as a quadratic equations in u, where u = e^x. Solve for u first then for x 1) e^2x - e^x =6 2)e^-2x - 3e^-x = -2

Question 785014: Given the equations, rewrite each as a quadratic equations in u, where u = e^x.
Solve for u first then for x
1) e^2x - e^x =6
2)e^-2x - 3e^-x = -2
Answer by fcabanski(1391) (Show Source): You can put this solution on YOUR website!
If u = then - when multiplying a base to a power by the same base to a different power, add the powers.

means

(u-3)(u+2) = 0

u = 3 and u = -2

Use the inverse function of e which is ln.

x = ln(3) = approximately 1.0986

For u=-2, do the same.

x = ln(-2): ln of a negative number is not a real number. So discard this answer. If you want to find the imaginary number solution, it's = approximately

Multiply all terms by to get

(2u-1)(u-1) = 0

2u = 1 so u = 1/2 so and thus x = ln(1/2) = approximately -.693

u=1 so so x = ln(1) = 0

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