# find all real numbers x for logx=x/100

• Apr 8th 2011, 05:08 PM
bigwave
find all real numbers x for logx=x/100
Find (with an error of less than $\displaystyle \frac{1}{10}$ ) all real numbers x for which $\displaystyle \log_{10}x = \frac{x}{100}$

I assume this can be rewritten as $\displaystyle x = 10^{\frac{x}{100}$ but don't know what next step would be

the answers given are 1.0238 and 237.57 which I got with the TI89 but not sure the steps taken to derive it.

thanks for help

r(Cool)
• Apr 8th 2011, 07:40 PM
Sambit
The only way is to use the method of iteration. I made up the following code in R:

[php]d=1.5;x=1.5
while(d>.000001)
{
x.updated=100*log(x,base=10)
d=abs(x.updated-x)
x=x.updated
}
x.updated[/php]

[php]
d=1.5;x=1.5
while(d>.000001)
{
x.updated=10^(x/100)
d=abs(x.updated-x)
x=x.updated
}
x.updated[/php]

Run it in R; you will get the results.
• Apr 12th 2011, 01:03 PM
bigwave
Quote:

Originally Posted by Sambit
The only way is to use the method of iteration

however this problem is out of book written in 1971 before writting a program for iteration was open to most students
I would presume there would be another way to solve it.

just curious
r
• Apr 12th 2011, 01:32 PM
SpringFan25
you can deduce from the shape of the graph that there are at most 2 solutions. (the right hand side is a straight line, the LHS has an always decreasing gradient).

I assume there is a clever way of getting the answer, but since i dont know it i would note that a talylor expansion of the LHS will give us a polynomial, which hopefully can be solved using conventional methods.

Try a taylor expansion about x=1 of $\displaystyle log_{10}x$

NB:
f'(x) = x^{-1}log_{10} e
f''(x) = -x^{-2}log_{10}e

Taylor expansion:
$\displaystyle log_{10}x \approx log_{10}1 + (x-1)x^{-1}k -0.5x^{-2}(x-1)^2k$

where k= log_10(e)

**thinking ahead, this is going to be a cubic. it's too late in the evening for that so ill use a 1 step taylor expansion instead

$\displaystyle log_{10}x \approx log_{10}1 + (x-1)x^{-1}k$

So solve
$\displaystyle log_{10}1 + (x-1)x^{-1}k = 0.01x$

$\displaystyle 0 + (x-1)x^{-1}k = 0.01x$

$\displaystyle (x-1)k = 0.01x^2$

This is a quadratic in x, with solutions

x=1.02 (pretty good!)
x=42 (not so good!)

presumably you could get closer by keeping more terms in the taylor expansion and solving a higher order polynomial.
• Apr 12th 2011, 02:41 PM
pickslides
Quote:

Originally Posted by bigwave
however this problem is out of book written in 1971 before writting a program for iteration was open to most students
I would presume there would be another way to solve it.

just curious
r

Well how about an iterative method by hand? Bisection method comes to mind.
• Apr 12th 2011, 10:16 PM
Sambit
Quote:

Originally Posted by bigwave
however this problem is out of book written in 1971 before writting a program for iteration was open to most students
I would presume there would be another way to solve it.

just curious
r

Of course you can do the method of iteration by hand. Method of bisection is a good idea, as pickslides said. I wrote that code for solving it with lesser labour and computation than by hand.
• Apr 12th 2011, 11:25 PM
CaptainBlack
Quote:

Originally Posted by bigwave
Find (with an error of less than $\displaystyle \frac{1}{10}$ ) all real numbers x for which $\displaystyle \log_{10}x = \frac{x}{100}$

I assume this can be rewritten as $\displaystyle x = 10^{\frac{x}{100}$ but don't know what next step would be

the answers given are 1.0238 and 237.57 which I got with the TI89 but not sure the steps taken to derive it.

thanks for help

r(Cool)

One step of Newton Raphson on $\displaystyle f(x)=x-10^{x/100}$ with $\displaystyle x_0=1$ will do the job (possibly another step to confirm the error is small enough).

CB
• Apr 13th 2011, 10:53 AM
bigwave
thanks everyone what i learned was quite valuable