## Compare big exponentials

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### Compare big exponentials

I stumbled across a problem a long time ago where I was given several exponencial numbers, and was my mission to find the largest. I'm trying to remember what is the best way to solve this... I think there was a property in logarithms that allowed this to be made quite fast, but can't seem to get it to work.

Example, from this list, which is the biggest:
2^3^4^5
6^1^3^6
9^3^2^2
8^8^8^8

Thanks
mrlinx
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### Re: Compare big exponentials

I think this should work.

Suppose we are comparing a^b^c and d^e^f,

assume d = a^x
.. x = log(d) / log(a)

and so d^e^f = (a^x)^e^f = a^(xe)^f
then we have to compare b^c with (xe)^f and see which one is bigger!! This can be found using the above method.

sohel
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### Re: Compare big exponentials

sohel wrote:and so d^e^f = (a^x)^e^f = a^(xe)^f

No, (a^x)^e^f = a^(x e^f)
(Remember that exponentiation is right-associative: a^b^c = a^(b^c).)
mf
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### Re: Compare big exponentials

Oops! You are, of course, right.
Then, is there any solution with logarithms, or do we have to bring number theoretic algorithms into action?

sohel
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