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6.6.72 log and e problem

karush

Well-known member
Jan 31, 2012
3,084
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
Mar 27, 2017
27
Newton's method solves this readily enough.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
854
Yes, either use a calculator or do a whole lot of arithmetic with paper and pencil!
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,460
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
Jan 31, 2012
3,084

karush

Well-known member
Jan 31, 2012
3,084
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
Jan 26, 2012
1,460
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

Well-known member
Jan 31, 2012
3,084

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
854
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!