n_kelthuzad said:
I was trying to understand superlog and superroot but I get only 3/4 of them. Can anyone just explain, in a non-textbook way, such that:
I can understand without any post-calc knowledge
http://en.wikipedia.org/wiki/Superlog
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or just explain:
c=x^x
how to find x?
or c=x^x^x^x^x^x^x....
thanks. just to say again that I am not even doing calc in school, but I am doing all maths myself. So will you explain without using any complex analysis and other things?
Hey n_kelthuzad and welcome to the forums.
The reason why we need complex analysis is because the solution x could actually be negative. When x is negative then we are looking at most likely a complex number solution depending on what x actually is.
So let's for the moment assume that x > 0. We only consider a finite number of exponentiations: Let's look at two first.
c = x^x^x. Assume x > 0 and c > 0 (It has to be otherwise it won't work).
log_c(c) = log_c(x^x^x) = 1
But log_c(x^x^x) = x^xlog_c(x) = 1.
This meanx x^x = 1/log_c(x).
Now let's do the same thing again:
log_c(x^x) = log_c(1/log_c(x)) = - log_c(log_c(x)) since log(1) = 0 and log(a/b) = log(a) - log(b).
But log_c(x^x) = xlog_c(x) which means:
xlog_c(x) = -log_c(log_c(x)) which means
x = -log_c(log_c(x)) / log_c(x)
To solve this, you need to use a computer for the general case. Now log_c(x) has to be greater than 0 otherwise you won't get a solution. This means that x > c for the real case (remember no complex numbers).
Since log_c(x) > 0 and since x > 0 this means log_c(log_c(x)) < 0 since we have a minus sign but this means log_c(log_c(x)) < 0 which means log_c(x) is in 0 < x < 1.
If those conditions are satisfied we use a computer program which takes in a formula and finds what is called the root of the equation: in this particular example - we have x + log_c(log_c(x))/log_c(x) = 0 and we have to solve for x. The computer will then do its magic and get an approximate answer for x.
Also just in case you are wondering log_c(x) = ln(x)/ln(c) where ln is the natural logarithm function which has a known formula.