My graphing calculator can’t display numbers higher than 10^100. In calculus, we were talking about finding the derivative of f(x)=x^x and I was curious to see what the graph looked like.
I played around with it and little and basically noticed that after a certain x-value, if I try to evaluate the function, the calculator will say “OVERFLOW” (if the evaluation is greater than 10^10).
Then I wanted to find out what the highest x-value is that I can plug in and still get an actual response from the calculator.
Of course, I was able to through guess and check, find out that it was around 56.961… (if I remember correctly), but I wanted to see how one could algebraically solve this to get an exact answer.
I set it up:
10^100=x^x
ln(10^100)=ln(x^x)
100ln(10)=xln(x)
There I am stuck (and I have tried rearranging it in all sorts of ways but none of them worked). Now I asked my teacher and he says the he doesn’t know of a way to solve that algebraically, but there might be one. Though he also says that not everything can be solved algebraically.
Do any of you know?
This isn’t a homework problem. This is pure curiosity.
Sometimes, transcendental equations may have closed form solutions, but that’s not common (and you don’t generally find such solutions algebraically). Usually, you need to use a numeric or graphical method to find solutions, as you have done.
You can solve it via Newton-Raphson iteration if you like. We want to find the zero of:
f(x) = xln(x) - 100ln(10)
f’(x) = ln(x) + 1
So we iterate:
x’ = x - (xln(x) - 100ln(10)) / (ln(x) + 1)
Start with a guess of 50.
x’ = 50 - (195.601 - 230.258) / (3.912 + 1) = 57.055
x’ = 57.055 - (230.734 - 230.258) / (4.044 + 1) = 56.961 (already at your answer)
This method will roughly double your significant digits at each step, so with just one or two more iterations you’ll reach the limits of your calculator.
Even when transcendental equations do have solutions, they’re generally of the form “Let’s create a new function that we’ll define as the solution to that particular problem”. Some problems come up repeatedly, so that those purpose-defined functions come in very useful, sometimes even to the point that we forget that they’re special functions. For instance, the problem f’’(x) = -f(x) shows up a lot, and so we invented the sine function to solve that problem. The problem you’re posing, though, is not one that comes up very often, so there might not be a suitable function for it, or if there is, it’s not well-known.