SOLUTION: i worked this problem like this:
e^2x - 3e^x + 2 = 0
e^2x - 3e^x = -2
e^(2x\x^3) = e^(-2)
(2\x^2)= e^(-2)
2= e^(-2x^2)
(2\e^-2) = x^2
sqrt(14.778)= sqrt(x^2)
x=3.844
i t
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-> SOLUTION: i worked this problem like this:
e^2x - 3e^x + 2 = 0
e^2x - 3e^x = -2
e^(2x\x^3) = e^(-2)
(2\x^2)= e^(-2)
2= e^(-2x^2)
(2\e^-2) = x^2
sqrt(14.778)= sqrt(x^2)
x=3.844
i t
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Question 209731: i worked this problem like this:
e^2x - 3e^x + 2 = 0
e^2x - 3e^x = -2
e^(2x\x^3) = e^(-2)
(2\x^2)= e^(-2)
2= e^(-2x^2)
(2\e^-2) = x^2
sqrt(14.778)= sqrt(x^2)
x=3.844
i tried my answer. but it doesnt work out.
what did i do wrong?
thanks in advance
You can put this solution on YOUR website!
Not wrong, but doesn't help.
e^((2x)/(x^3)) = e^(-2)
???! There is no way to arrive at this statement from the previous one.
The key to solving the equation, , is to recognize that
the equation can then be written as:
the equation is therefore a quadratic equation in
If you have trouble seeing the last two, we can use a substitution to make it clearer. Let . Replacing with q we get: .
We can solve this for q (aka ) with the quadratic formula or by factoring:
In order for this product to be zero, one of the factors must be zero: or
Solving these we get: or
Of course we are not interested in "q". We are interested in "x". So we can substitute back in for "q": or
To solve these for x we will find the natural log of each side: or
Using one of the properties of logarithms: we get:
x*ln(e) = ln(2) or x*ln(e) = ln(1)
Since the ln(e) is 1 by definition and ln(1) is zero:
x = ln(2) or x = 0
(