# How can I solve this equation analytically? **URL:** \ **Category:** Factual Questions\ **Created:** [April 8, 2011, 11:48am UTC](https://boards.straightdope.com/t/how-can-i-solve-this-equation-analytically/577647 "2011-04-08T11:48:27Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [April 8, 2011, 11:48am UTC](https://boards.straightdope.com/t/how-can-i-solve-this-equation-analytically/577647/1 "2011-04-08T11:48:27Z")
This is not homework, I am not enrolled at any school. I am trying to brush up on my Math skills and I have come accross a problem that I should know how to work out. The equation is e^x - (10/7)x - 1 = 0 Using Newton’s method I have detemrined that the two solutions are 0 and 0.675471592932108. How can I get that by manipulating the equation? ---
**Author:** ![Half\_Man\_Half\_Wit](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/half_man_half_wit/32/21766_2.png) [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)\ **Post date:** [April 8, 2011, 12:08pm UTC](https://boards.straightdope.com/t/how-can-i-solve-this-equation-analytically/577647/2 "2011-04-08T12:08:23Z")
There’s no closed-form solution in terms of elementary functions. You can solve it in terms of the Lambert W-function, but you’re probably better off doing it numerically. Or just ask [Wolfram Alpha](http://www.wolframalpha.com/input/?i=e%5Ex+-+10%2F7*x+-+1+%3D+0)… ---
**Author:** ![Saffer](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saffer/32/6440_2.png) [@Saffer](https://boards.straightdope.com/u/Saffer)\ **Post date:** [April 8, 2011, 12:27pm UTC](https://boards.straightdope.com/t/how-can-i-solve-this-equation-analytically/577647/3 "2011-04-08T12:27:37Z")
Thanks, I have a long way to go before I will be casually using the “Lambert W-function”! ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [April 8, 2011, 12:54pm UTC](https://boards.straightdope.com/t/how-can-i-solve-this-equation-analytically/577647/4 "2011-04-08T12:54:51Z")
You can re-arrange the equation and take the natural logarithm of each side, giving **x = ln(ax + 1)** where, in your case, **a = 10/7**. If you expand the right hand side, you get ( see [Natural logarithm - Wikipedia](http://en.wikipedia.org/wiki/Natural_logarithm) , or a book of tables or infinite series) **ln(ax + 1 ) = ax - ((ax)[sup]2[/sup])/2 + ((ax)[sup]3[/sup])/3 - …** Equating this to **x** , you immediately see that \*\* x = 0\*\* is one solution. You can get your other solution by taking successively more terms of the expansion. Note that you must have \*\* |ax| \< 1 \*\*, which you do in both cases. That’s not closed-form for the second solution, but I don’t think you’re going to find one. ---
**Author:** ![Manlob](https://avatars.discourse-cdn.com/v4/letter/m/96bed5/32.png) [@Manlob](https://boards.straightdope.com/u/Manlob)\ **Post date:** [April 13, 2011, 11:17pm UTC](https://boards.straightdope.com/t/how-can-i-solve-this-equation-analytically/577647/5 "2011-04-13T23:17:44Z")
Writing in terms of parameter “a”: e^x - a\*x - 1 = 0, a good approximation can be obtained by expressing in terms of the “a”, although I only did it terms of b = 1-a. x\_0=b\*(12_b^3 - 100_b^2 +180_b -90) / (2_b^4 - 32_b^3 + 105_b^2 - 120\*b +45) Inserting b=-3/7 (a=10/7) gives a result within 4e-6 of the exact value. Using a higher order version of Newton’s method allows that result to be improved: Let g=1-(1+a_x\_0)/exp(x\_0) and t=1-a/exp(x\_0), then a better approximation is: x=x\_0 + 3_g\*(g-2_t^2)/(6_t^3-6_t_g+g^2) This gives a results within 8e-22 of the exact solution. Not exact, but pretty close.