SOLUTION: I need help with the problem {{{e^(2x+1)=9e^(1-x)}}} The question says to solve for the missing variable. SO what i did was made both sides in terms of e and then set the exponents

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Question 210767This question is from textbook Single Variable Calculus
: I need help with the problem e%5E%282x%2B1%29=9e%5E%281-x%29 The question says to solve for the missing variable. SO what i did was made both sides in terms of e and then set the exponents equal to each other and i get 8/11 which i am told is not correct. help please? This question is from textbook Single Variable Calculus

Found 2 solutions by stanbon, tutor_paul:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
e^(2x+1)=9e^(1-x)
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Take the natural log of both sides:
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2x+1 = ln(9) + (1-x)
3x = ln(9)
x = [ln(9)]/3
x = 0.7324..
Cheers,
Stan H.

Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
e%5E%282x%2B1%29=9e%5E%281-x%29
Take the natural log of both sides:
ln%28e%5E%282x%2B1%29%29=ln%289%2Ae%5E%281-x%29%29
use properties of logarithms:
2x%2B1=ln%289%29%2B1-x
solve for x:
3x=ln%289%29
highlight%28x=%28ln%289%29%29%2F3%29
============
Good Luck,
tutor_paul@yahoo.com