Question 499302: How to solve |3*e^(2x)-5*e^(x)|<2 ?
Answer by chessace(471)   (Show Source):
You can put this solution on YOUR website! Not too easily.
First note that e^2x = (e^x)^2 and use y = e^x
Then have |3y^2 - 5y| < 2
Then break into the usual 2 cases for absolute value:
3y^2 - 5y < 2 and 3y^2 - 5y > -2
Then find zeroes after sending over the +-2:
3y^2 - 5y -2 = 0 and 3y^2 - 5y + 2 = 0
These factor into
(3y+1)(y-2)=0 and (3y-2)(y-1)=0
With roots
-1/3, 2 and 2/3, 1
By considering large (+ and -) values for y, it's clear that the function
f(y) = 3y^2 - 5y as y increases in value,
is > 2 until = 2 at y = -1/3
[then passes through the origin]
is = -2 at y = 2/3 and is < -2 until
[it reaches its minimum of - 2.833 at y = 5/6]
is = -2 again at y = 1,
and inceases to 2 at y = 2, then is > 2.
So the valid intervals for y are
(-1/3, 2/3) and (1, 2).
Now we need to relate this to x, where y = e^x.
So take natural logs of all 4 endpoints.
The -1/3 is artificial, e^x is never < 0.
Also e^0 = 1, so only need to look up 2 logs.
Final answer: x must belong to interval (-infinity, -0.405465) or
the interval (0 , 0.693147)
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