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

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

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

I assume this can be rewritten as $x = 10^{\frac{x}{100}$ 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

2. 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.

3. 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

4. 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 $log_{10}x$

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

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

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

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

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

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

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

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.

5. 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.

6. 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.

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

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

CB

8. thanks everyone what i learned was quite valuable