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How do you solve $- 3 e^{2 x} + 16 = 5$ $-3e^(2x) + 16=5$ and find any extraneous solutions?

1 Answer

Apr 12, 2017

$x = \frac{ln (\frac{11}{3})}{2}$ $x=ln(11/3)/2$
$XXX$ $color(white)("XXX")$ ...maybe I missed something, but I did not find any extraneous solutions.

Explanation:

$- 3 e^{2 x} + 16 = 5$ $-3e^(2x)+16=5$
$XXX \to - 3 e^{2 x} = - 11$ $color(white)("XXX")rarr -3e^(2x)=-11$

$XXX \to e^{2 x} = \frac{11}{3}$ $color(white)("XXX")rarr e^(2x)=11/3$

$XXX \to 2 x = ln (\frac{11}{3})$ $color(white)("XXX")rarr2x=ln(11/3)$

$XXX \to x = \frac{ln (\frac{11}{3})}{2} \approx 0.64964$ $color(white)("XXX")rarr x=ln(11/3)/2 ~~0.64964$ (I used a calculator for this approximation)

Checking this solution back into the original equation demonstrates that the solution is not extraneous. (Again, I used a calculator for this).

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