How can I solve the problem of $e^x-\ln{x}=4$ without a calculator?

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In summary: 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{
  • #1
karush
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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
 
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  • #2
Newton's method solves this readily enough.
 
  • #3
Yes, either use a calculator or do a whole lot of arithmetic with paper and pencil!
 
  • #4
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.
 
  • #5
romsek said:
Newton's method solves this readily enough.
don't think I would know how to use Newtons method on this
 
  • #7
Prove It said:
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
 
  • #8
karush said:
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...
 
  • #9
Prove It said:
Which is a CAS...

yes but its not a handheld that requires e charging and is easily lost
and a hard to read screen
 
  • #10
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!
 
  • #11
Mahalo that was a great help
 

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