# Thread: Numerical Methods

1. ## Numerical Methods

quick question
how do i find the number of real solutions for this:

e^x = 1 / x

i assume i would need to take logs of both sides so i can evaluate the x on the left
if i take logs of the LHS, I end up with just x, how do I take logs of the other side?

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Keep in mins it's not log base ten it's natural log (LN). There is no numerical answer because it is a limit. You can rewrite the right side as ln(1)/ln(x) and the ln(1)=0. The ln(infinity)=infinity. I guess you could say zero.
Last edited by goatboy; 01-24-2005 at 01:11 PM.

3. the book says the answers 1

4. Well, we know that log(xy)=log(x)+log(y), and log(x/y)=log(x)-log(y), so by taking the natural logarithm of both sides, we get: x = ln(1) - ln(x) --> x = -ln(x) --> x + ln(x) = 0

I don't know how to solve it from there. Maybe taking the derivative of both sides would work.. but I don't remember how to do that for logarithms!

In any case, the answer is not 1, as you can confirm by replacing x by 1 in the equation.

5. Maybe there's an easier method or some rule I don't remember anymore, but this is what I came up with:

Rewrite your expression as x * e^x - 1. Now, the problem consists of finding how many intercepts with the ordinate axis the equation y = x * e^x - 1 has.
After derivating and equaling to zero:

e^x * (x + 1) = 0

Which means, "extrema" are to be found in x = -1 and x = -infinity (last one is of course no proper extremum). The second derivative is

e^x * (x + 2)

which, evaluated for x = -1 yields e^x, which is positive regardless of x. That means, the extremum in x = -1 is a minimum.

Now we have to examine the limit of y for x -> -infinity. We find that to be -1 (after transforming a little bit and applying l'Hopital's rule).

To sum up, the function comes from x = -infinity (where it "has" a value of -1) and has only one extremum (a minimum) in x = -1. That means, it doesn't intercept the ordinate axis between x = -infinity and x = -1. Since there aren't any other extrema and at x = -1 we are at a minimum, the function must grow continually from x = -1, which means it will only intercept the ordinate axis in one point, which is the only real root of the expression. So, the answer is 1 real solution.

Demondo: feel free to point out any eventual mistakes I made here...

P.S. If you take ln at both sides, the right one would be (ln1 - lnx), that is, -lnx.

6. Why not just do the simplest method and just sketch the functions... it is really easy to see they only intercept once. As for another approach, here's the one I'd use:

x is either: > 0 , < 0 , = 0 So...
For x < 0 ; e^x > 1 and 1/x < 0 so there will be no intercept there.
For x = 0 ; e^x = 1 and 1/x <is undefined so that doesn't work.
So now you just need to look at the case where x > 0:

(1/x)' = -1/x^2, which is negative for any positive value of x (other than 0)... thus it is ALWAYS decreasing. (Read: 1/x > 1/(x+Epsilon))

(e^x)' = e^x, which is positive for any real value of x. Thus it is always increasing. (Read: e^x < e^(x+Epsilon))

Epsilon > 0

----------------

Once you get the primary behavior of the functions down, you just need to see if they cross or not in the domain: ]0,infinity[. To do this look at four limits:

lim e^x = 1
x -> 0 (from the positives)

lim 1/x = infinity
x -> 0 (from the positives)

Therefore when you are close to 0, e^x < 1/x

lim e^x = infinity
x -> infinity

lim 1/x = 0
x -> infinity

Therefore when you are in higher values e^x > 1/x

By continuity of the functions in the interval ]0,infinity[ there must be ONE point where they cross. (Note: there can't be more than one, because we looked at all three cases x > 0 , x = 0 , and x < 0. AND we showed that the each function is either strictly increasing OR strictly decreasing in the domain where they intersect... so they can't turn around and intersect a second time.)

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