x^(x^x) - (x^x)^x = 0
Thus, x^(x^x) = (x^x)^x
Let's open it up by taking log on both sides...
(x^x)*log(x) = x* log(x^x)
(x^x)*log(x) = x*x*log(x)
If x==1 equation is satisfied as log(x) becomes 0..
so x=1 is definitely a solution. what if when x != 1
cancelling log(x) on both the sides..
x^x =
how many values of x are possible in the following equation.
x^(x^x) - (x^x)^x = 0
where a^b =power(a,b).
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2 solutions - {1,2}
On Fri, Jul 29, 2011 at 11:10 AM, vaibhav_iiit honeys...@gmail.com wrote:
how many values of x are possible in the following equation.
x^(x^x) - (x^x)^x = 0
where a^b =power(a,b).
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