# Stuck at solving equation

#### Trigger12

Hi.

I can't find an answer to the equation $$\displaystyle e^x+x-3=0$$. I'm not sure how to solve it.

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#### SDK

Just apply Newton's method. This gives a solution $x = .7921...$. Since the expression is monotone increasing, this must be a unique solution.

#### Trigger12

I have already done that in the second question where they tell you to use NewtonÂ´s method. But the first question is: Show that we can find the value of a, solving the equation e^x+x-3=0.

#### Trigger12

The tangent to the curve $$\displaystyle y=e^{2-x}$$ in $$\displaystyle x=2$$ cuts the curve $$\displaystyle y=e^x$$ in a point where x=a.

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#### mathman

Forum Staff
The equation has no analytic solution, only numerical.

#### Trigger12

Ok, but how to I proceed to solve it numerical?

#### JeffM1

I have already done that in the second question where they tell you to use NewtonÂ´s method. But the first question is: Show that we can find the value of a, solving the equation e^x+x-3=0.
Have you thought about applying the mean value theorem to the function

$f(x) = e^x + x - 3.$

Is f(x) continuous?

What is the sign of f(0)?

What is the sign of f(1)?

What does all that imply about the existence of a such that f(a) = 0?

#### Trigger12

Have you thought about applying the mean value theorem to the function

$f(x) = e^x + x - 3.$

Is f(x) continuous?

What is the sign of f(0)?

What is the sign of f(1)?

What does all that imply about the existence of a such that f(a) = 0?
Ahh! I didn't think about solving it that way.

#### JeffM1

Ahh! I didn't think about solving it that way.
Without understanding the exact language of the problem, I cannot be sure my suggestion is even partially relevant. And my Swedish is non-existent so you are on your own in terms of judging how helpful the suggestion is.

#### Trigger12

Without understanding the exact language of the problem, I cannot be sure my suggestion is even partially relevant. And my Swedish is non-existent so you are on your own in terms of judging how helpful the suggestion is.
In the second question, I used Newton's method to find an approximate value of a, and got 0,792059, that is correct. My problem is that they want us to find the same value, but by solving the equation given over in the first question.

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