# 6.6.72 log and e problem

#### karush

##### Well-known member
solve $e^x-\ln{x}=4\quad x\approx1.48 \quad x\approx 005$

ok I could only do this with a calculator but need steps

Last edited:

#### romsek

##### Member
Newton's method solves this readily enough.

• karush

#### Country Boy

##### Well-known member
MHB Math Helper
Yes, either use a calculator or do a whole lot of arithmetic with paper and pencil!

#### Prove It

##### Well-known member
MHB Math Helper
To answer your question, even though it's easy to show that a root exists, it's actually impossible to get an exact form for.

As others have suggested, if you want to get an approximate solution, you'll need an iterative method such as Newton's Method, or get the CAS to solve it for you.

#### karush

##### Well-known member
Newton's method solves this readily enough.
don't think I would know how to use Newtons method on this

#### karush

##### Well-known member
To answer your question, even though it's easy to show that a root exists, it's actually impossible to get an exact form for.

As others have suggested, if you want to get an approximate solution, you'll need an iterative method such as Newton's Method, or get the CAS to solve it for you.
I quit buying handheld calculators I had a TI inspire CS CAS but the keyboard wore out

W|A can calculate it

#### Prove It

##### Well-known member
MHB Math Helper
I quit buying handheld calculators I had a TI inspire CS CAS but the keyboard wore out

W|A can calculate it
Which is a CAS...

• karush

#### karush

##### Well-known member
Which is a CAS...
yes but its not a handheld that requires e charging and is easily lost
and a hard to read screen

#### Country Boy

##### Well-known member
MHB Math Helper
To use Newton's method, we select some starting point, hopefully close to a root of the equation but not necessarily. If that point happens to be a root, we are done. If not then we construct the tangent line to the graph of the function at that point and solve the linear equation to see where that tangent line crosses the x-axis. Hopefully, if that new x value is not a root, it is closer so we do it again.

Specifically, if the problem is to solve f(x)= 0 and we choose $x_0$ as our starting point, the tangent line is $y= f'(x_0)(x- x_0)+ f(x_0)$. Setting that equal to 0, $f'(x_0)(x- x_0)+ f(x_0)= 0$ so $f'(x_0)(x- x_0)= -f(x_0)$, $x- x_0= -\frac{f(x_0)}{f'(x_0)}$ and then $x= x_0- \frac{f(x_0)}{f'(x_0)}$.

THAT is "Newton's method"-to solve f(x)= 0, select some starting $x_0$ and calculate $x_1= x_0- \frac{f(x_0)}{f'(x_0)}$. Then calculate $x_2= x_1- \frac{f(x_1)}{f'(x_1)}$ and keep repeating that until you get the accuracy you want. Typically you know you are close to the desired root because the values are close together. For example, if you want the root "correct to three decimal places" keep repeating until successive values of x are correct to three decimal places.

Here, $f(x)= e^x- log(x)- 4$ and $f'(x)= e^x-\frac{1}{x}$ so that $x_1= x_0- \frac{f(x_0)}{f'(x_0)}= x_0-\frac{e^{x_0}- log(x_0)- 4}{e^{x_0}- \frac{1}{x_0}}$.
Choose some starting value, say $x_0=1$ or $x_0= 2$ and start calculating!

If you were as old as I am you would have learned to use tables to find the exponential and logarithm but for God's sake use a calculator!

• karush and topsquark