st (1) m = root(n), if n<0 then root(n) is undefined.
given: m,n>0; m^n < n^m?
st(1) m=n^1/2,
LHS n^(n/2)
RHSn^(n^1/2) --->
LHS n/2
RHSn^1/2,
LHS n
RHS 2*n^1/2 Not sufficient, as plug in n=1 and n=9 gives two inequality conditions;
st(2) n>5 alone is Not Sufficient, as we are missing the relationship data for n and m.
Combined st(1&2) Sufficient as
LHS m^n is always greater than
RHS n^m
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:yellowho wrote:Ran across this on a problem X^(Y^4). Is there a way to reduce this or a different to rewrite this?
It can be written as X^(Y^2)^2. What's the problem?
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