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?
Reducing Exponents
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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?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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yellowho
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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?
[quote="Anurag@Gurome"][quote="yellowho"]Ran across this on a problem X^(Y^4). Is there a way to reduce this or a different to rewrite this?[/quote]
It can be written as X^(Y^2)^2. What's the problem?[/quote]
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?
[quote="Anurag@Gurome"][quote="yellowho"]Ran across this on a problem X^(Y^4). Is there a way to reduce this or a different to rewrite this?[/quote]
It can be written as X^(Y^2)^2. What's the problem?[/quote]
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Night reader
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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
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: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?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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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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