GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 03 Jul 2020, 12:12 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # NEW!!! Tough and tricky exponents and roots questions

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 64939
NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

47
1
204 00:00

Difficulty:

(N/A)

Question Stats: 91% (01:45) correct 9% (00:26) wrong based on 347 sessions

### HideShow timer Statistics

Exponents and roots problems are very common on the GMAT. So, it's extremely important to know how to manipulate them, how to factor out, take roots, multiply, divide, etc. Below are 11 problems to test your skills. Please post your thought process/solutions along with the answers.

I'll post OA's with detailed solutions tomorrow. Good luck.

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) $$x^2-y^2=23$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029239

2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ an integer?
(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.
(2) $$\sqrt{z}$$ is not an integer.

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029240

3. If $$x>y>0$$ then what is the value of $$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$?
(1) $$x+y=4+2\sqrt{xy}$$
(2) $$x-y=9$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029241

4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029242

5. If $$x$$ and $$y$$ are negative integers, then what is the value of $$xy$$?
(1) $$x^y=\frac{1}{81}$$
(2) $$y^x=-\frac{1}{64}$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029243

6. If $$x>{0}$$ then what is the value of $$y^x$$?
(1) $$\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}$$
(2) $$x\neq{1}$$ and $$x^y=1$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967.html#p1029244

7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*x}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029245

8. What is the value of $$x^2+y^3$$?
(1) $$x^6+y^9=0$$
(2) $$27^{x^2}=\frac{3}{3^{3y^2+1}}$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029246

9. If $$x$$, $$y$$ and $$z$$ are non-zero numbers, what is the value of $$\frac{x^3+y^3+z^3}{xyz}$$?
(1) $$xyz=-6$$
(2) $$x+y+z=0$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029247

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) $$2^x+3^y=\sqrt[(x+y)]{25}$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029248

11. What is the value of $$xy$$?
(1) $$3^x*5^y=75$$
(2) $$3^{(x-1)(y-2)}=1$$

Solution: tough-and-tricky-exponents-and-roots-questions-125967-20.html#p1029249
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

22
57
2. If x, y, and z are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ an integer?
(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.
(2) $$\sqrt{z}$$ is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> $$xyz=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube --> if $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3=odd \ perfect \ cube$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3=odd \ perfect \ cube$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (2) $$\sqrt{z}\neq{integer}$$ then $$z\neq{1}$$, therefore it must be $$3^3$$ (from 1) --> $$x$$ is a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

_________________
##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

THEORY TO TACKLE THE PROBLEMS ABOVE:
For more on number theory check the Number Theory Chapter of Math Book: math-number-theory-88376.html

EXPONENTS

Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number $$a$$ multiplied $$n$$ times can be written as $$a^n$$, where $$a$$ represents the base, the number that is multiplied by itself $$n$$ times and $$n$$ represents the exponent. The exponent indicates how many times to multiple the base, $$a$$, by itself.

Exponents one and zero:
$$a^0=1$$ Any nonzero number to the power of 0 is 1.
For example: $$5^0=1$$ and $$(-3)^0=1$$
• Note: the case of 0^0 is not tested on the GMAT.

$$a^1=a$$ Any number to the power 1 is itself.

Powers of zero:
If the exponent is positive, the power of zero is zero: $$0^n = 0$$, where $$n > 0$$.

If the exponent is negative, the power of zero ($$0^n$$, where $$n < 0$$) is undefined, because division by zero is implied.

Powers of one:
$$1^n=1$$ The integer powers of one are one.

Negative powers:
$$a^{-n}=\frac{1}{a^n}$$

Powers of minus one:
If n is an even integer, then $$(-1)^n=1$$.

If n is an odd integer, then $$(-1)^n =-1$$.

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
$$a^n*b^n=(ab)^n$$

$$\frac{a^n}{b^n}=(\frac{a}{b})^n$$

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

$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$ (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
$$a^n*a^m=a^{n+m}$$

$$\frac{a^n}{a^m}=a^{n-m}$$

Fraction as power:
$$a^{\frac{1}{n}}=\sqrt[n]{a}$$

$$a^{\frac{m}{n}}=\sqrt[n]{a^m}$$

ROOTS

Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.

General rules:
• $$\sqrt{x}\sqrt{y}=\sqrt{xy}$$ and $$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$$.

• $$(\sqrt{x})^n=\sqrt{x^n}$$

• $$x^{\frac{1}{n}}=\sqrt[n]{x}$$

• $$x^{\frac{n}{m}}=\sqrt[m]{x^n}$$

• $${\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}$$

• $$\sqrt{x^2}=|x|$$, when $$x\leq{0}$$, then $$\sqrt{x^2}=-x$$ and when $$x\geq{0}$$, then $$\sqrt{x^2}=x$$

• When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

• Odd roots will have the same sign as the base of the root. For example, $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$.
_________________
Intern  Joined: 12 Oct 2011
Posts: 17
Location: United States
GMAT 1: 720 Q50 V36
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
1
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

10<y<x<13
y = 11
x = 12

Sufficient

(2) x^2-y^2=23
(x+y)(x-y) = 23
x-y=1
x+y=23
x=12
y=11

Sufficient

Option
Intern  Joined: 12 Oct 2011
Posts: 17
Location: United States
GMAT 1: 720 Q50 V36
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer?
(1) y is an even perfect square and z is an odd perfect cube.
(2) \sqrt{z} is not an integer.

2700 = 2*2*3*3*3*5*5
(1) y is an even perfect square and z is an odd perfect cube.

y = 100 or 4
z = 27
x = 1 or 25

Yes, √x is an integer.
Sufficient

(2) \sqrt{z} is not an integer.

Insuff

Option
GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2276
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

3
1
Bunuel wrote:

4. If $$xyz\neq{0}$$ is $$x^{-4}*\sqrt{y}*z^{-2}<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

6. If $$x\neq{0}$$ then what is the value of $$y^x$$?
(1) $$\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}$$
(2) $$x\neq{1}$$ and $$x^y=1$$

7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*y}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

9. If $$\frac{x}{y^{-3}}+\frac{y}{x^{-3}}=\frac{1}{(\sqrt{2}xy)^{-2}}$$, then what is the value of $$xy$$?
(1) $$x^2=y^2$$
(2) $$x^3>y^3$$

At a quick glance, a few comments - hopefully I haven't made any errors:

The meaning of Q4 would be more clear if the terms were enclosed in brackets (right now the asterisk symbol appears to be part of an exponent): 4. $$\text{If } xyz\neq{0} \text{ is } \left( x^{-4} \right) \left( \sqrt{y} \right) \left( z^{-2} \right) <0 \text{ ?}$$

Q6 is a bit problematic. If y turns out to be 0, we need to know that x is positive for the expression in question to be defined. I don't know if, in Statement 2, you had in mind the 'trap' that x might be -1, but for Statement 1 to work, that possibility needs to be ruled out in advance.

In Q7, I imagine you meant to write 'x' instead of 'y' in Statement 1, since there's no other mention of y anywhere.

There seems to be something wrong with Q9. If both statements are true, x and y need to have opposite signs, but that would make the left side of the equation in the stem negative and the right side positive, which is clearly impossible. The question also needs to rule out the possibility that x or y are equal to 0, since that would make terms in the equation in the stem undefined (you can't raise zero to a negative power).
_________________
GMAT Tutor in Montreal

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2276
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
1
rijul007 wrote:
2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer?
(1) y is an even perfect square and z is an odd perfect cube.
(2) \sqrt{z} is not an integer.

2700 = 2*2*3*3*3*5*5
(1) y is an even perfect square and z is an odd perfect cube.

y = 100 or 4
z = 27
x = 1 or 25

Yes, √x is an integer.
Sufficient

There's another possibility here. It is possible that z = 1. In that case, x could be, say, 3^3, and then its square root would not be an integer.
_________________
GMAT Tutor in Montreal

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

IanStewart wrote:
Bunuel wrote:

4. If $$xyz\neq{0}$$ is $$x^{-4}*\sqrt{y}*z^{-2}<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

6. If $$x\neq{0}$$ then what is the value of $$y^x$$?
(1) $$\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}$$
(2) $$x\neq{1}$$ and $$x^y=1$$

7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*y}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

9. If $$\frac{x}{y^{-3}}+\frac{y}{x^{-3}}=\frac{1}{(\sqrt{2}xy)^{-2}}$$, then what is the value of $$xy$$?
(1) $$x^2=y^2$$
(2) $$x^3>y^3$$

At a quick glance, a few comments - hopefully I haven't made any errors:

The meaning of Q4 would be more clear if the terms were enclosed in brackets (right now the asterisk symbol appears to be part of an exponent): 4. $$\text{If } xyz\neq{0} \text{ is } \left( x^{-4} \right) \left( \sqrt{y} \right) \left( z^{-2} \right) <0 \text{ ?}$$

Q6 is a bit problematic. If y turns out to be 0, we need to know that x is positive for the expression in question to be defined. I don't know if, in Statement 2, you had in mind the 'trap' that x might be -1, but for Statement 1 to work, that possibility needs to be ruled out in advance.

In Q7, I imagine you meant to write 'x' instead of 'y' in Statement 1, since there's no other mention of y anywhere.

There seems to be something wrong with Q9. If both statements are true, x and y need to have opposite signs, but that would make the left side of the equation in the stem negative and the right side positive, which is clearly impossible. The question also needs to rule out the possibility that x or y are equal to 0, since that would make terms in the equation in the stem undefined (you can't raise zero to a negative power).

For Q4: enclosed the terms in brackets to avoid confusion.
For Q6: yes, there is a typo, in the stem it should read $$x>0$$ instead of $$x\neq{0}$$.
For Q7: yes, there is a typo, should be x instead of y.
For Q9: yes, I took the stem from one question (not finished yet) and the statements from another. Already substituted this question.
_________________
Intern  Joined: 12 Oct 2011
Posts: 17
Location: United States
GMAT 1: 720 Q50 V36
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

2
IanStewart wrote:
rijul007 wrote:
2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer?
(1) y is an even perfect square and z is an odd perfect cube.
(2) \sqrt{z} is not an integer.

2700 = 2*2*3*3*3*5*5
(1) y is an even perfect square and z is an odd perfect cube.

y = 100 or 4
z = 27
x = 1 or 25

Yes, √x is an integer.
Sufficient

There's another possibility here. It is possible that z = 1. In that case, x could be, say, 3^3, and then its square root would not be an integer.

Oops.. Thanks for pointing out Ian.
Manager  Joined: 23 Oct 2011
Posts: 77
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

2
1
Bunuel wrote:
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

if we can find the sign of y we can answer the question.

(1) $$\sqrt{y}>\sqrt{x^2}>0$$ since it is$$[\sqrt{x^2}]=[\sqrt[]{x}]$$

Sufficient

(2) $$y^3>\frac{1}{z{^4}}>0$$ since $$z^4>0$$

Sufficient

Therefore, D
Manager  Joined: 23 Oct 2011
Posts: 77
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
2
Bunuel wrote:
5. If $$x$$ and $$y$$ are negative integers, then what is the value of $$xy$$?
(1) $$x^y=\frac{1}{81}$$
(2) $$y^x=-\frac{1}{64}$$

(1) y must be even for the statement to be true (only then will the result be positive). $$1/81=$$ 3^(-4) ---> the pairs to make 4 in order for x and y to be integers is 1*4 or 2*2. Therefore the only way to construct it is if y=-2 or y=-4.

if x=-3 and y=-4 --->$$x^y=\frac{1}{81}$$ ---> x*y=12
if x=-9 and y=-2 --->$$x^y=\frac{1}{81}$$ ---->x*y=18

Insufficient

(2) x must be odd for the statement to be True (only then will the result be negative).$$-1/64=-2^6$$ ---> the pairs to make 6 in order for x and y to be integers are 1*6 or 2*3. Therefore the only way to construct it is if x=-1 or x=-3.

if y=-4 and x=-3 ---> $$y^x=-\frac{1}{64}$$ --->x*y=12
if y=-64 and x=-1 ---> $$y^x=-\frac{1}{64}$$ ---> x*y=64

Insufficient.

(1)+(2)---> x*y=12 ---> C
Manager  Joined: 23 Oct 2011
Posts: 77
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

2
1
Bunuel wrote:
7. If $$x$$ is a positive integer is $$\sqrt{x}$$ an integer?
(1) $$\sqrt{7*x}$$ is an integer
(2) $$\sqrt{9*x}$$ is not an integer

(1) x must have a 7 raised at an odd power as a factor. It could also have any number raised to an even power as a factor ---> therefore $$\sqrt{x}$$ will never be an integer because of the 7. Sufficient

(2) $$\sqrt{9*x}$$=$$3\sqrt{x}$$---> since 3 is an integer, $$\sqrt{x}$$ is an not an integer.

Sufficient

Therefore, D
Intern  Joined: 13 Jul 2011
Posts: 10
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1
1-D
2-D
3-E
4-D
5-B
6-D
7-D
8-E
9-B
10-D
11-A

Intern  Joined: 22 Aug 2011
Posts: 2
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

1-D
2-C
3-C
4-D
5-C
6-B
7-D
8-D
9-C
10-E
11-A

May you please post the OA GMAT Tutor P
Joined: 24 Jun 2008
Posts: 2276
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

7
2
Bunuel wrote:

3. If $$x>y>0$$ then what is the value of $$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$?
(1) $$x+y=4+2\sqrt{xy}$$
(2) $$x-y=9$$

Since in both of the posts above, the answer to this question was given incorrectly, I thought I'd post a quick solution. Using the difference of squares,

$$x - y = (\sqrt{x})^2 - (\sqrt{y})^2 = (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$$

So we can simplify the question by using this factorization in the denominator:

$$\frac{\sqrt{2x} + \sqrt{2y}}{x-y} = \frac{\sqrt{2} (\sqrt{x} + \sqrt{y})}{(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) } = \frac{\sqrt{2}}{\sqrt{x} - \sqrt{y}}$$

So if we can find the value of $$\sqrt{x} - \sqrt{y}$$, we can answer the question.

Now from Statement 1, we have

\begin{align} x + y &= 4 + 2\sqrt{xy} \\ x - 2\sqrt{xy} + y &= 4 \\ (\sqrt{x} - \sqrt{y})^2 &= 4 \\ \sqrt{x} - \sqrt{y} &= 2 \end{align}
(here we know the root is 2, and not -2, since x > y). So Statement 1 is sufficient.
_________________
GMAT Tutor in Montreal

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

19
25
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

14
45
3. If $$x>y>0$$ then what is the value of $$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$?
(1) $$x+y=4+2\sqrt{xy}$$
(2) $$x-y=9$$

$$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$ --> factor out $$\sqrt{2}$$ from the nominator and apply $$a^2-b^2=(a-b)(a+b)$$ to the expression in the denominator: $$\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}$$. So we should find the value of $$\sqrt{x}-\sqrt{y}$$.

(1) $$x+y=4+2\sqrt{xy}$$ --> $$x-2\sqrt{xy}+y=4$$ --> $$(\sqrt{x}-\sqrt{y})^2=4$$ --> $$\sqrt{x}-\sqrt{y}=2$$ (note that since $$x-y>0$$ then the second solution $$\sqrt{x}-\sqrt{y}=-2$$ is not valid). Sufficient.

(2) $$x-y=9$$. Not sufficient.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

9
19
4. If $$xyz\neq{0}$$ is $$(x^{-4})*(\sqrt{y})*(z^{-2})<0$$?
(1) $$\sqrt{y}>\sqrt{x^2}$$
(2) $$y^3>\frac{1}{z{^4}}$$

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

(1) $$\sqrt{y}>\sqrt{x^2}$$ --> as even root from positive number ($$x^2$$ in our case) is positive then $$\sqrt{y}>\sqrt{x^2}>0$$, (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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

9
21
5. If $$x$$ and $$y$$ are negative integers, then what is the value of $$xy$$?
(1) $$x^y=\frac{1}{81}$$
(2) $$y^x=-\frac{1}{64}$$

(1) $$x^y=\frac{1}{81}$$ --> as both $$x$$ and $$y$$ are negative integers then $$x^y=\frac{1}{81}=(-9)^{-2}=(-3)^{-4}$$ --> $$xy=18$$ or $$xy=12$$. Note that as negative integer (x) in negative integer power (y) gives positive number (1/81) then the power must be negative even number. Not sufficient.

(2) $$y^x=-\frac{1}{64}$$ --> as the result is negative then $$x$$ must be negative odd number --> $$y^x=-\frac{1}{64}=(-4)^{-3}=(-64)^{-1}$$ --> $$xy=12$$ or $$xy=64$$. Not sufficient.

(1)+(2) Only one pair of negative integers $$x$$ and $$y$$ satisfies both statements $$x=-3$$ and $$y=-4$$ --> $$xy=12$$. Sufficient.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 64939
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

### Show Tags

5
19
6. If $$x>0$$ then what is the value of $$y^x$$?
(1) $$\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}$$
(2) $$x\neq{1}$$ and $$x^y=1$$

(1) $$\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}$$ --> $$4^{(x+y)^2-(x-y)^2}=128^{xy}$$ --> applying $$a^2-b^2=(a-b)(a+b)$$ we'll get: $$4^{4xy}=128^{xy}$$ --> $$2^{8xy}=2^{7xy}$$ --> $$8xy=7xy$$ --> $$xy=0$$, since given that $$x>0$$ then $$y=0$$ hence $$y^x=0^x=0$$. Sufficient.

(2) $$x\neq{1}$$ and $$x^y=1$$ --> since $$x>0$$ and $$x\neq{1}$$ then the only case $$x^y=1$$ to hold true is when $$y=0$$ --> $$y^x=0^x=0$$. Sufficient.

_________________ Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 14 Jan 2012, 14:43

Go to page    1   2   3   4    Next  [ 70 posts ]

# NEW!!! Tough and tricky exponents and roots questions  