# Musings concerning maximal expressions leading to a logarithmic equation

• Apr 24th 2012, 02:41 PM
NowIsForever
Musings concerning maximal expressions leading to a logarithmic equation
Upon awakening this morning I had this idea about the largest expression that could be formed from a fixed collection of digits, all the same number, using standard operations such as multiplication, exponentiation, etc.

For example suppose, the collection is 2,2,2,2. My first thought was that 2^2^2^2 would be the answer for the maximum value, however $22^2^2$ is larger, and $2^2^2^2$ is larger still.

Then I noticed that $22^2^2$ is closer to $2^2^2^2$ (by ratio) than $33^3^3$ is to $3^3^3^3$. The situation is further divergent as the value of the digit climbs to 9.

Define d*d as 10·d+d and d*d*d as d·100+d·10+d etc., where d is not only allowed to be a digit, but any real number. (Numbers less than or equal to zero, although allowed, may or may not be reasonable).

So this thought formed: what would be the value of d such that $(d*d)^(^d^*^d^)$ = $d^(^d^*^d^*^d^)$?

It would be a solution to the equation: $(10x+x) log_x (10x+x)= 100x+10x+x$.

Not wishing to look for an analytic solution to the equation prior to determining a numerical result; I proceeded to do that obtaining

x = 1.3018267624863938102032556203448

using my Windows calculator after 39 iterations (I know, I could have written a C program to do this in less than half the time, but I'm on vacation for three weeks, I like doing it, and I probably would not have discovered the following result had I written the program):

I noticed this as a convergent during my calculations: $log_x$ (10x+x) = 111/11. (Rofl)

It is not obvious to me why this should be so; perhaps it would fall out of an analytic solution quite easily, but I have yet to look for one, and I'm posting this so that all and sundry may have a shot at it as I work on it myself.

Any feedback would be greatly appreciated.
• Apr 24th 2012, 02:53 PM
ignite
Re: Musings concerning maximal expressions leading to a logarithmic equation
$x=11^\frac{11}{100}$ solves the given equation.