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It can be written as X^(Y^2)^2. What's the problem?yellowho wrote:Ran across this on a problem X^(Y^4). Is there a way to reduce this or a different to rewrite this?
unless you use differentials from analytic algebra, there's no way to reduce this expression (X^(Y^4))`=4X^(3Y^3)yellowho wrote:Ran across this on a problem X^(Y^4). Is there a way to reduce this or a different to rewrite this?
yellowho wrote:Thanks anurag. Actually the problem just makes you compare two numbers that are both raise to some power. One way to evaluate is to reduce or just compare the exponents if you already know the base. I was just looking for an alternative method.
If m and n are positive integers, is m^n < n^m?
(1) m = root(n)
(2) n > 5
What if we don't know whether N is positive? Then you have to evaluate 3 cases Positive, Zero, and Negative right? Would you eliminate negative outright because it would involve imaginary number? So you would just deal with positive and zero case?
Anurag@Gurome wrote:It can be written as X^(Y^2)^2. What's the problem?yellowho wrote:Ran across this on a problem X^(Y^4). Is there a way to reduce this or a different to rewrite this?
Yes, we can easily eliminate the negative integers as the first statement involves a square root: n can't be negative as it is under the square root and m can't be negative as we consider only the positive values of a square root. Hence, we have to deal with positive and zero case.yellowho wrote:What if we don't know whether N is positive? Then you have to evaluate 3 cases Positive, Zero, and Negative right? Would you eliminate negative outright because it would involve imaginary number? So you would just deal with positive and zero case?
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