# How do you solve e^x + lnx = 5?

May 9, 2016

1.522. Let $f \left(x\right) = {e}^{x} + \ln x - 5$. Locate the root in (n, n+1) by sign test f(n)f(n+1)<0. Bisect the interval and choose the half in which f passes the sign test. Halving continues, for desired accuracy...

#### Explanation:

The root could not be got in mathematical exactitude. Only an approximation to any sd-accuracy could be obtained. Graphical methods lack accuracy. Iterative methods are fine.

Bisection method is explained here. This is quite easy for anyone. There are other faster numerical methods..

Let $f \left(x\right) = {e}^{x} + \ln x - 5$.

As $f \left(1\right) f \left(2\right) = \left(- 2.28 . .\right) \left(3.08 . .\right) < 0$, a root lies between 1 and 2.

Now $f \left(1.5\right) = - 0.11 . . < 0$ and quite small in magnitude.

Use your discretion for a short enclosure.

Choose (1.5, 1.6) for sign test, and so on, until you get the desired accuracy.

I find f(1.5215)f(1.5220) < 0. So rounded 4-sd approximation of the root is 1.522..

This algorithms for the bisection method and its variations are programmable, for any befitting computer language...