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

1 Answer
May 9, 2016

1.522. Let #f(x)=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...


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(x)=e^x+ln x-5#.

As #f(1)f(2) = (-2.28..)(3.08..) < 0#, a root lies between 1 and 2.

Now #f(1.5) = -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...