Futility Closet posted the other day that log 237.5812087593 = 2.375812087593.

Make a formal statement of the problem that 2.375812087593 solves and show that there’s exactly one other solution.

**Related posts**:

Futility Closet posted the other day that log 237.5812087593 = 2.375812087593.

Make a formal statement of the problem that 2.375812087593 solves and show that there’s exactly one other solution.

**Related posts**:

x/log(x)=100

And the other solution is one very close to 1 from infinitity; that’s another number such that its logarithm is 1/100 of its value. Below 1 it would be negative, and zero and below it would not be defined. (I’m using non-calculus high school math here.)

The basic equation is log(100x) = x. This simplifies to log(100) + log(x) = x, and thus 2 + log(x) – x = 0.

Let y = 2 + log(x) – x. We are looking for the places where this function intercepts the x axis. x=2.3758… is one of them. This function is undefined for x0, y approaches negative infinity.

Take the first derivative, and you get y’ = (1/x) – 1. When x1, it’s negative, and the slope is decreasing.

So, y starts at negative infinity when x is close to 0, increases until x=1, then decreases after that. Since one x-intercept is at a value of x that is greater than the point at which the function is maximized, there must be some point 0<x<1 at which the function intersects the x-axis. This will be the only other solution. It's close to x=0.01024.

solve[10^x-100*x=0]

{{x -> 0.010238552281443276}, {x -> 2.375812087593426}}

log(100x)=x, which translates to Wok’s equation. I used my TI-83 to find the intersection of y=100x and y=10^x.

x = log(100*x)

-> x = log(100) + log(x)

-> x = log(x) + 2

This gives the second solution as being slightly above 0.01. There cannot be more than two solutions to this equation because the left-hand side is a straight line and the right-hand side is strictly concave everywhere.

Not my most elegant proof, but it’s been a long time since I’ve done this.

The problem is this: log(x) = x/100

1) Both log(x) and x/100 are continuous over the set of all positive numbers.

2) Log(x) is a curve that always has a negative (and decreasing) curvature.

3) because curvature always has the same sign, a straight line can only intersect it once, twice, or not at all.

4) Because x/100 is a straight line, it may not intersect log(x) more than twice

5) for values of x=0 and x=infinity, x/100 is greater than log(x)

6) For some values of x (example 10), x/100 is less than log(x)

7) From this we can conclude that

a) The x/100 and log(x) must intersect

b) they must intersect twice.

So, if 2.3…. is a solution, there is exactly one other solution.

Surprised nobody gave the link to Wolfram Alpha. The graph helps to illustrate.

Bonus puzzle: For what k does x = log10(x) + k have one and only one solution?

http://www.wolframalpha.com/input/?i=log%28100x%29+%3D+x&a=*FunClash.log-_*Log10.Log-

For Nemo’s bonus puzzle, the answer is

1/ln(10)+log10(ln(10))

which is about 0.79651017.

Jonathan —

Yup. Although sticking with the base-10 logs, I would write it log(e) – log(log(e)).

The number 2.375812087593 is the root of the following equation:

x^2- 3.375812087593x + 2.375812087593 = 0

The only other solution appears to be 1. What do I win?

Yes, that’s a nicer way to write it. I just wrote down what I got from solving it, which gave me that ugly mess of logs of different bases…

Fascinating, did no one else immediately see the generalization to

x = log x + n, wherenis any integer? Obviously that doesn’t have just one other solution, but it makes a very neat pattern. There are two series of solutions, one series which quickly approaches 1, and one which grows a little faster thann. The easiest way to see the whole thing is to look at the graph ofx – log(x).Or exp(x)/x : http://goo.gl/YyZI0