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 \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(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
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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 \(\sqrt[4]{x}\), 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, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
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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

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.
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Oops.. Thanks for pointing out Ian.

(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

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

please post the OA.

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.
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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.

Answer: A.
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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.

Answer: C.
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