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NEW!!! Tough and tricky exponents and roots questions

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New post 23 Jul 2014, 09:29
Kconfused wrote:
Bunuel wrote:
1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?
(1) \(100<y^2<x^2<169\)
(2) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.


Hello Bunuel, for the second statement, you seem to indicate that since the difference of the squares of two positive integers is a prime number, the two integers are consecutive on the number scale. Does this always hold?
Thank you!


Yes, positive integer solutions of x^2-y^2=prime must be consecutive, because of x-y=1 (x=y+1).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 21 Aug 2014, 03:30
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.
Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.
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New post 21 Aug 2014, 04:14
joshnsit wrote:
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.
Hi Bunuel, I have a minor doubt. Using the corollary a^3+b^3=(a+b)(a^2-a*b+b^2) and plugging in a=x^2 and b=y^3 in statement 1,
we can say that (x^2)^3+(y^3)^3=(x^2+y^3)(x^4-x^2y^3+y^6)=0
Since any of expressions (x^2+y^3) or (x^4-x^2y^3+y^6) can be zero. It seems that statement 1 is insufficient by this logic.
Kindly comment.


The point is that only real roots of x^4 - x^2y^3 + y^6 = 0 is x = y = 0, for which x^2 + y^3 = 0.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 09 Jan 2015, 05:28
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.


Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.
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New post 09 Jan 2015, 05:39
aniteshgmat1101 wrote:
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.


Hi Bunuel,
I have a confusion. The problem statement with option no. 1 can be written as follows:
if a^3 + b^3 = 0 (1st option) then what is the value of a + b ? (where a = x^2 and b = y^3).
Now, we know that (a^3 + b^3) = (a+b) * (a^2 - a*b + b^2). (a^3 + b^3) has 2 factors.
And according to the explanation, since, (a^3 + b^3) =0 => a^3 = -(b^3) => a = -b => a + b = 0. So, whenever a^3 + b^3 is 0, a+b will also be 0 !!!!
Can these steps be true always? Please reply... i am really confused.


Yes, if a^3 + b^3 = 0, then a + b = 0.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 12 Jul 2015, 16:28
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.



Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 12 Jul 2015, 16:59
1
ranaazad wrote:
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.




Hello sir,
It was not mentioned in the question stem whether x, y and z are integers. In this question, if x, y and z were non-integers, wouldn't the procedure be different? Would you please explain?

Thanks! :D


This question has a trick in that it will be true for both integers and non-integers alike:

1. x^2 > 0 no matter x=integer or not. Similarly, \(\sqrt[4]{x^2}\) will be positive for x=integer or non-integer. Try it out. Let x = 1 , \(\sqrt[4]{1^2} = 1 > 0\) and if x= 0.5, \(\sqrt[4]{0.5^2} = 0.707 >0.\)

Thus it does not matter whether you take x = integer or a fraction.

Coming back to the question,

Per statement 1, as mentioned above, \(\sqrt[4]{1^2} = 1 > 0\) ---> \(\sqrt[5]{y} > 0\) -----> y >0. Thus the statement \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\) will be false. Thus this statement is sufficient.

Per statement 2, z^4 > 0 whether z =1 or z= 0.5 ----> 1/(z^4) >0 ----> y^3 >0---> y>0 (satisfies for both, y =1 or y = 0.25). Thus the given statement \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\) will be false. Thus this statement is sufficient.

As both the given statements are sufficient, the correct answer is D.

whenever you are in doubt, always test the given statements/conditions with different values (integers, fractions , etc). This way you will get a better picture.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 12 Jul 2015, 17:03
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.



Hello Bunuel,
Nice explanations. I can't figure out why exponents over exponents act different sometimes. When to multiply and when to add them? Here, why isn't (x^2)^3 = x^8 and (-y^3)^3 = -y^27, please explain? As we know, 3^3^3 = 27.

Could you please explicate?
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New post 13 Jul 2015, 00:56
1
ranaazad wrote:
Bunuel wrote:
8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). Sufficient.

(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\) --> \(3^{3x^2}=\frac{3}{3^{3y^2}*3}\) --> \(3^{3x^2}*3^{3y^2}=1\) --> \(3^{3x^2+3y^2}=1\) --> \(3x^2+3y^2=0\) (the power of 3 must be zero in order this equation to hold true) --> \(x^2+y^2=0\) the sum of two non-negative values is zero --> both \(x\) and \(y\) must be zero --> \(x=y=0\) --> \(x^2+y^3=0\). Sufficient.

Answer: D.



Hello Bunuel,
Nice explanations. I can't figure out why exponents over exponents act different sometimes. When to multiply and when to add them? Here, why isn't (x^2)^3 = x^8 and (-y^3)^3 = -y^27, please explain? As we know, 3^3^3 = 27.

Could you please explicate?


Please read the second post of this topic: new-tough-and-tricky-exponents-and-roots-questions-125967.html#p1027927

\((a^m)^n=a^{mn}\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 14 Oct 2016, 13:21
Bunuel wrote:
10. If \(x\) and \(y\) are non-negative integers and \(x+y>0\) is \((x+y)^{xy}\) an even integer?
(1) \(2^{x-y}=\sqrt[(x+y)]{16}\)
(2) \(2^x+3^y=\sqrt[(x+y)]{25}\)

(1) \(2^{x-y}=\sqrt[(x+y)]{16}\) --> \(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\) --> equate the powers: \(x-y=\frac{4}{x+y}\) --> \((x-y)(x+y)=4\).

Since both \(x\) and \(y\) are integers (and \(x+y>0\)) then \(x-y=2\) and \(x+y=2\) --> \(x=2\) and \(y=0\) --> \((x+y)^{xy}=2^0=1=odd\), so the answer to the question is No. Sufficient. (Note that \(x-y=1\) and \(x+y=4\) --> \(x=2.5\) and \(y=1.5\) is not a valid scenario (solution) as both unknowns must be integers)

(2) \(2^x+3^y=\sqrt[(x+y)]{25}\) --> obviously \(\sqrt[(x+y)]{25}\) must be an integer (since \(2^x+3^y=integer\)) and as \(x+y=integer\) then the only solution is \(\sqrt[(x+y)]{25}=\sqrt[2]{25}=5\) --> \(x+y=2\). So, \(2^x+3^y=5\) --> two scenarios are possible:
A. \(x=2\) and \(y=0\) (notice that \(x+y=2\) holds true) --> \(2^x+3^y=2^2+3^0=5\), and in this case: \((x+y)^{xy}=2^0=1=odd\);
B. \(x=1\) and \(y=1\) (notice that \(x+y=2\) holds true) --> \(2^x+3^y=2^1+3^1=5\), and in this case: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A.



if y = 0 then statement 1 become 2^x = 2^2x , i.e. x = 2x , i.e. x = 0 not 2 , what am i doing wrong here if you plz??
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 15 Oct 2016, 02:49
yezz wrote:
Bunuel wrote:
10. If \(x\) and \(y\) are non-negative integers and \(x+y>0\) is \((x+y)^{xy}\) an even integer?
(1) \(2^{x-y}=\sqrt[(x+y)]{16}\)
(2) \(2^x+3^y=\sqrt[(x+y)]{25}\)

(1) \(2^{x-y}=\sqrt[(x+y)]{16}\) --> \(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\) --> equate the powers: \(x-y=\frac{4}{x+y}\) --> \((x-y)(x+y)=4\).

Since both \(x\) and \(y\) are integers (and \(x+y>0\)) then \(x-y=2\) and \(x+y=2\) --> \(x=2\) and \(y=0\) --> \((x+y)^{xy}=2^0=1=odd\), so the answer to the question is No. Sufficient. (Note that \(x-y=1\) and \(x+y=4\) --> \(x=2.5\) and \(y=1.5\) is not a valid scenario (solution) as both unknowns must be integers)

(2) \(2^x+3^y=\sqrt[(x+y)]{25}\) --> obviously \(\sqrt[(x+y)]{25}\) must be an integer (since \(2^x+3^y=integer\)) and as \(x+y=integer\) then the only solution is \(\sqrt[(x+y)]{25}=\sqrt[2]{25}=5\) --> \(x+y=2\). So, \(2^x+3^y=5\) --> two scenarios are possible:
A. \(x=2\) and \(y=0\) (notice that \(x+y=2\) holds true) --> \(2^x+3^y=2^2+3^0=5\), and in this case: \((x+y)^{xy}=2^0=1=odd\);
B. \(x=1\) and \(y=1\) (notice that \(x+y=2\) holds true) --> \(2^x+3^y=2^1+3^1=5\), and in this case: \((x+y)^{xy}=2^1=2=even\).

Two different answers. Not sufficient.

Answer: A.



if y = 0 then statement 1 become 2^x = 2^2x , i.e. x = 2x , i.e. x = 0 not 2 , what am i doing wrong here if you plz??


If y=0, then we'll have \(2^{x}=\sqrt[x]{16}\):

\(2^x=2^{\frac{4}{x}}\);

\(x=\frac{4}{x}\);

\(x^2=4\)

\(x=2\) or \(x=-2\) (discard since x is non-negative).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 13 Nov 2017, 11:06
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.


I could not understand the highlighted part. I also don't understand when to consider both positive and negative solutions and when to consider only positive solution when dealing with roots. Please explain.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 13 Nov 2017, 21:46
1
scrumptious829 wrote:
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.


I could not understand the highlighted part. I also don't understand when to consider both positive and negative solutions and when to consider only positive solution when dealing with roots. Please explain.


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 16 Feb 2018, 10:38
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.



Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 16 Feb 2018, 11:19
1
duahsolo wrote:
Bunuel wrote:
4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.



Hi Bunuel,

Can you kindly help me understand why 4th root of x^2 > 0 and not 4th root of x^2 = 4th root of |x| similar to square of x^2 = square root of |x| as shown in the step 2 of your solution to the question @ https://gmatclub.com/forum/what-is-x-2- ... 99157.html?

Thank you


Even roots from a non-negative number is non-negative, so 0 or positive: \(\sqrt[even]{something}\geq 0\), not matter what that "something" is (provided it's positive or 0 because even roots from negative numbers are not defined for the GMAT) )

Next, \(\sqrt[4]{x^2}=\sqrt{\sqrt{x^2}}=\sqrt{|x|}\), which must be positive, so (1) basically says that \(\sqrt[5]{y}>0\), or that y > 0.

Finally, if it were
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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