## GRE Exponents

We are absolutely sure that you will get problem on GRE exponents on test day. In our chapters on GRE exponents we have covered from each and every property to difficult problems on GRE exponents. If you go through our article carefully problems can’ t trick you if they meant to trick you.

Problem 1) $${ (\frac { 3x }{ 4 } ) }^{ 2 }\quad +\quad { (\frac { { y }^{ 2 } }{ 2 } ) }^{ 4 }$$

a) $$(\frac { 9{ x }^{ 2 }+{ y }^{ 8 } }{ 4 } )$$

b) $$(\frac { 9{ x }^{ 2 }+{ y }^{ 8 } }{ 16 } )$$

c) $$(\frac { 9{ x }^{ 2 }+{ y }^{ 4 } }{ 16 } )$$

d) $$(\frac { 3{ x }^{ 2 }+{ y }^{ 8 } }{ 16 } )$$

Sol. $$\frac { 9{ x }^{ 2 } }{ 16 } +\frac { { y }^{ 8 } }{ 16 } \\ =(\frac { 9{ x }^{ 2 }+{ y }^{ 8 } }{ 16 } )$$

(b) is correct.

Problem 2)  $$(\frac { { x }^{ 4 }{ y }^{ -2 } }{ { x }^{ -3 }{ y }^{ 4 } } )$$

a)  $$\frac { { x }^{ 7 } }{ { y }^{ 4 } }$$

b)  $$\frac { { x }^{ 7 } }{ { y }^{ 6 } }$$

c)  $$\frac { { x }^{ 3 } }{ { y }^{ 6 } }$$

d)  $$\frac { { x }^{ 4 } }{ { y }^{ 6 } }$$

Sol.  $$(\frac { { x }^{ 4 }{ y }^{ -2 } }{ { { x }^{ -3 }{ y }^{ 4 } } } )\\ =\quad (\frac { { x }^{ 4 }{ x }^{ 3 } }{ { y }^{ 6 } } )\\ =\quad (\frac { { x }^{ 7 } }{ { y }^{ 6 } } )$$

(b) is correct.

Problem 3) $$(\frac { { r }^{ 0 }{ s }^{ 4 } }{ t } )\div { (\frac { 3s }{ t } ) }^{ 2 }$$

a) $$(\frac { 3{ s }^{ 2 }t }{ 9 } )$$

b) $$(\frac { { s }^{ 2 }t }{ 9 } )$$

c) $$(\frac { { s }^{ 6 } }{ { t }^{ 3 } } )$$

d) $$(\frac { { 3s }^{ 6 } }{ { t }^{ 3 } } )$$

Sol. $$(\frac { { r }^{ 0 }{ s }^{ 4 } }{ t } )\div (\frac { 9{ s }^{ 2 } }{ { t }^{ 2 } } )\\ =(\frac { { s }^{ 4 } }{ t } )(\frac { { t }^{ 2 } }{ 9{ s }^{ 2 } } )\\ =(\frac { 1 }{ 9 } )({ s }^{ 2 }t)$$

(b) is correct.

Problem 4) $${ (\frac { -3 }{ { x }^{ 2 } } ) }^{ 3 }{ (6xy) }^{ 0 }(\frac { { x }^{ 5 } }{ 9 } )$$

a) $$-3{ x }^{ 3 }$$

b) $$\frac { 3 }{ x }$$

c) $$\frac { -3 }{ x }$$

d) $$3{ x }^{ 3 }$$

Sol. $$(\frac { -27 }{ { x }^{ 6 } } )(\frac { { x }^{ 5 } }{ 9 } )\\ =\quad (\frac { -3{ x }^{ 3 } }{ { x }^{ 4 } } )\\ =\quad (\frac { -3 }{ x } )$$

(c) is correct.

Problem 5) 80 is divisible by $${ 2 }^{ x }$$

Quantity A:       x

Quantity B:       3

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol. 80 is divisible by $${ 2 }^{ 4 }$$ & therefore also by

$${ 2 }^{ 3 }{ ,\quad 2 }^{ 2 }\quad { ,2 }^{ 1 }$$ and $${ 2 }^{ 0 }$$.

Thus x could be 0, 1, 2, 3 or 4 & could therefore be less than, equal to or greater than 3.

Thus relationship cannot be determined.

Problem 6) For which of the following positive integers is the square of the integer divided by the cube root of the same integer equal to nine times that integer?

a) 4         b) 8            c)16            d) 27          e) 125

Sol.   The correct answer is 27.

Let integer be a

$${ \frac { { a }^{ 2 } }{ { a }^{ \frac { 1 }{ 3 } } } \quad =\quad 9a\\ \\ { a }^{ 2-\frac { 1 }{ 3 } -1 }\quad =\quad { 3 }^{ 2 } }\\ { a }^{ \frac { 2 }{ 3 } }\quad =\quad { 3 }^{ 2 }\\ a\quad =\quad { 3 }^{ 3 }\quad =\quad { 27 }^{ Ans }$$

(d) is correct.

Problem 7) If $${ 2 }^{ k }-{ 2 }^{ k+1 }+{ 2 }^{ k-1 }\quad =\quad { 2 }^{ k }m$$, what is the value of m?

a) -1

b) $$-\frac { 1 }{ 2 }$$

c) $$\frac { 1 }{ 2 }$$

d) 1

e) 2

Sol.  $${ 2 }^{ k }-{ 2 }^{ k+1 }+{ 2 }^{ k-1 }={ 2 }^{ k }m\\ { 2 }^{ k }(1-2+\frac { 1 }{ 2 } )={ 2 }^{ k }m\\ m\quad =\quad (-1+\frac { 1 }{ 2 } )\quad =\quad -\frac { 1 }{ 2 }$$

(b) is correct.

Problem 8) If the hash marks above are equally spaced, What is the value of p?

a) $$\frac { 3 }{ 2 }$$

b) $$\frac { 3 }{ 2 }$$

c) $$\frac { 3 }{ 2 }$$

d) $$\frac { 3 }{ 2 }$$

e) $$\frac { 512 }{ 125 }$$

Sol.

$$(\frac { 2 }{ 5 } )(4)\quad =\quad { p }^{ \frac { 1 }{ 3 } }$$

To determine the distance between Harsh marks, divide 2 (the distance from 0 to 2) by 5 (the number of strength the number line has been divided into).

The result is $$\frac { 2 }{ 5 }$$.

Therefore,

$$(\frac { 2 }{ 5 } )(4)\quad =\quad { p }^{ \frac { 1 }{ 3 } }\\ (\frac { 8 }{ 5 } )\quad =\quad { p }^{ \frac { 1 }{ 3 } }\\ p\quad =\quad { (\frac { 8 }{ 5 } ) }^{ 3 }\quad =\quad (\frac { 512 }{ 125 } )$$

(e) is correct.

Problem 9) What is the greatest prime factor of $${ 2 }^{ 99 }-{ 2 }^{ 96 }$$  ?

Sol.

$${ 2 }^{ 99 }-{ 2 }^{ 96 }={ 2 }^{ 96 }(8-1)={ 2 }^{ 96 }(7)\quad$$

Greatest prime factor = 7      Ans.

Problem 10) Which of the following is equal to

$$(\frac { { 10 }^{ -8 }{ 25 }^{ 7 }{ 2 }^{ 16 } }{ { 20 }^{ 6 }{ 8 }^{ -1 } } )\quad$$ ?

a) $$(\frac { 1 }{ 5 } )\quad$$

b) $$(\frac { 1 }{ 2 } )\quad$$

c) 2

d) 5

e) 10

Sol.

$$(\frac { { 10 }^{ -8 }{ 25 }^{ 7 }{ 2 }^{ 16 } }{ { 20 }^{ 6 }{ 8 }^{ -1 } } )\quad =\quad (\frac { { 5 }^{ -8 }{ 2 }^{ -8 }{ 5 }^{ 14 }{ 2 }^{ 16 } }{ { 2 }^{ 12 }{ 5 }^{ 6 }{ 2 }^{ -3 } } )\\ =\quad { 5 }^{ -8+14-6 }{ 2 }^{ 16-8+3-12 }\\ =\quad { 2 }^{ 19-20 }\\ =\quad { 2 }^{ -1 }\\ =\quad \frac { 1 }{ 2 }$$

(b) is correct.

Problem 11) Solve $$\sqrt { 4+\sqrt { 131+\sqrt { 154+\sqrt { 225 } } } }$$ ?

Sol.

$$\sqrt { 4+\sqrt { 131+\sqrt { 154+\sqrt { 225 } } } } \\ =\quad \sqrt { 4+\sqrt { 131+\sqrt { 154+15 } } } \\ =\quad \sqrt { 4+\sqrt { 131+13 } } \\ =\quad \sqrt { 4+\sqrt { 144 } } \\ =\quad \sqrt { 4+12 } \\ =\quad { 4 }^{ Ans }$$