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SSAT Upper Level Math : Properties of Exponents

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Example Questions

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Example Question #121 :How To Find The Properties Of An Exponent

What is $23\times23\times23\times23\times23\times23$ in exponential notation?

Possible Answers:

23^6

$6\times23$

$23\times6$

$6^{23}$

$6\times6\times6\times6\times6\times6\times6\times6\times6\times6\times6$

Correct answer:

23^6

Explanation:

Exponential notation includes a base number and an exponent. The base number is the number that is being multiplied and the exponent is how many times the base number is being multiplied to itself.

In this case,is our base number and it's being multiplied to itselftimes, so that is our exponent.

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Example Question #122 :How To Find The Properties Of An Exponent

$x^{5} = - 77$

y = -2

$(xy) ^{5} =?$

Possible Answers:

- 2,464

2,464

Correct answer:

2,464

Explanation:

Apply the Power of a Product Principle:

$(xy) ^{5} = x^{5} y ^{5}$

Setting $x^{5} = - 77$ and y = -2 , keeping in mind that an odd power of a negative number is negative:

$x^{5} y ^{5}$

$= -77 \cdot (-2 ) ^{5}$

$= -77 \cdot[ - (2 ^{5} )]$

$= -77 \cdot (-32)$

$= + (77 \cdot 32)$

= 2,464

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Example Question #123 :How To Find The Properties Of An Exponent

y > 0 and z> 0

$y^{2} = 12$

$z^{2} = 5$

Evaluate $(y z)^{3}$ .

Possible Answers:

$12\sqrt{30}$

$12\sqrt{15}$

$120 \sqrt{30 }$

$120 \sqrt{15}$

Correct answer:

$120 \sqrt{15}$

Explanation:

$y^{2} = 12$ andis positive, so

$y = \sqrt{12}$ .

The greatest perfect square factor of 12 is 4, so the radical can be simplified:

$y = \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}$

$z^{2} = 5$ , andis positive, so $z = \sqrt{5}$

By the Power of a Power Property,

$(y z)^{3}$

$= y^{3} z^{3}$

It is easiest to note that this can be broken up by the Product of Powers Principle, and evaluated by substitution:

$y^{3} z^{3}$

$= y^{2+ 1} z^{2+ 1}$

$= y^{2 } \cdot y \cdot z^{2 } \cdot z$

$=12 \cdot \sqrt{12} \cdot 5 \cdot \sqrt{5}$

$=12 \cdot 5 \cdot \sqrt{12} \cdot \sqrt{5}$

$=60 \cdot \sqrt{60}$

The greatest perfect square factor of 60 is 4, so the radical can be simplified:

$60 \cdot \sqrt{60}$

$=60 \cdot \sqrt{4} \cdot \sqrt{15}$

$=60 \cdot 2 \cdot \sqrt{15}$

$=120 \sqrt{15}$

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Example Question #124 :How To Find The Properties Of An Exponent

$x^{2} = 75$

$y^{2} = 3$

andare both positive.

Evaluate (4x +y) (4x -y) .

Possible Answers:

183

297

1,083

1,197

Correct answer:

1,197

Explanation:

By the difference of squares pattern:

(4x +y) (4x -y)

$= (4x) ^{2} - y^{2}$

By the Power of a Power Principle,

$(4x) ^{2} - y^{2}$

$= 4^{2}x^{2} - y^{2}$

$=16 x^{2} - y^{2}$

Substituting 75 and 3 for $x^{2}$ and $y^{2}$ , respectively:

=16 (75)- 3

=1,200- 3

=1,197

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Example Question #125 :How To Find The Properties Of An Exponent

$y ^{5} = -7$

$y^{20} = ?$

Possible Answers:

2,401

-2,401

Correct answer:

2,401

Explanation:

By the Power of a Power Principle,

$(y ^{5} ) ^{4} = y ^{5 \cdot 4} = y ^{20}$

Substitutingfor $y ^{5}$ , keeping in mind that an even power of any number must be positive:

$y ^{20} = (y ^{5} ) ^{4} =( -7)^{4} = 7 ^{4} = 7 \cdot 7 \cdot 7 \cdot 7= 2,401$

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Example Question #181 :Algebra

x> 0 and y > 0

$x^{2} = 8$

$y^{2} = 18$

Evaluate $(x+ y) ^{2}$ .

Possible Answers:

144

Correct answer:

Explanation:

By the perfect square trinomial pattern,

$(x+ y) ^{2}$

$=x^{2} +2xy+ y ^{2}$

$x^{2} = 8$ and $y^{2} = 18$ .

Also, by the Power of a Power Principle,

$(xy) ^{2} = x^{2} y^{2} = 8 \cdot 18 = 144$ ,

so, sinceandare both positive,

$xy = \sqrt{(xy) ^{2} }= \sqrt{144}= 12$ .

Therefore,

$(x+ y) ^{2}$

$=x^{2} +2xy+ y ^{2}$

= 8 +2 (12)+18

= 8 +24 +18

= 50

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Example Question #182 :Algebra

$y^{3} = -17$

$y^{9} = ?$

Possible Answers:

-4,913

4,913

Correct answer:

-4,913

Explanation:

By the Power of a Power Principle,

$(y ^{3} ) ^{3} = y ^{3 \cdot 3} = y ^{9}$

Therefore, we substitute, keeping in mind that an odd power of a negative number is also negative:

$y ^{9} = (y ^{3} ) ^{3}$

$= (-17) ^{3}$

$= - 17 ^{3}$

$= - (17 \cdot 17\cdot 17)$

= -4,913

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Example Question #183 :Algebra

x> 0 and y > 0

$x^{2} = 18$

$y^{2} = 8$

Evaluate $(x- y) ^{2}$ .

Possible Answers:

Correct answer:

Explanation:

By the perfect square trinomial pattern,

$(x- y) ^{2}$

$=x^{2} -2xy+ y ^{2}$

$x^{2} =1 8$ and $y^{2} = 8$ .

Also, by the Power of a Power Principle,

$(xy) ^{2} = x^{2} y^{2} = 18 \cdot 8 = 144$ ,

so, sinceandare both positive,

$xy = \sqrt{(xy) ^{2} }= \sqrt{144}= 12$ .

Therefore,

$(x- y) ^{2}$

$=x^{2} -2xy+ y ^{2}$

And, substituting:

$x^{2} -2xy+ y ^{2}$

=18 -2 (12)+8

=18 -24 +8

= 2

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Example Question #184 :Algebra

$t^{6} = 144$

$x^{6}= 81$

t > 0, x< 0

Evaluate the expression $(t-x) (t^{2}+tx+x^{2})$ .

Possible Answers:

Correct answer:

Explanation:

Multiply out the expression by using multiple distributions and collecting like terms:

$(t-x) (t^{2}+tx+x^{2})$

$= t (t^{2}+tx+x^{2}) - x (t^{2}+tx+x^{2})$

$= t \cdot t^{2}+ t \cdot tx+ t \cdot x^{2} - x \cdot t^{2}- x \cdot tx- x \cdot x^{2}$

$= t^{3}+ t^{2} x+ t x^{2} - t^{2} x- tx^{2} - x^{3}$

$= t^{3}+ t^{2} x- t^{2} x + t x^{2} - tx^{2} - x^{3}$

$= t^{3} - x^{3}$

Since $(t^{3} ) ^{2} = t ^{3 \cdot 2 } = t^{6}$ by the Power of a Power Principle,

$t^{3} = \pm \sqrt{ t^{6}} = \pm \sqrt{144} = \pm 12$ .

However,is positive, so $t^{3}$ is as well, so we choose $t^{3} = 12$ .

Similarly,

$x^{3} = \pm \sqrt{x^{6}} = \pm \sqrt{81} = \pm 9$ .

However, sinceis negative, as an odd power of a negative number, $x^{3}$ is as well, so we choose $x^{3} = -9$ .

Therefore, substituting:

$(t-x) (t^{2}+tx+x^{2}) = t^{3} - x^{3} = 12 - (-9) = 21$

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Example Question #121 :Properties Of Exponents

andare both positive integers; A is odd. What can you say about the number

$(A + B + 1)^{A+B+1}$ ?

Possible Answers:

$(A + B + 1)^{A+B+1}$ is even ifis odd, and odd ifis even.

$(A + B + 1)^{A+B+1}$ is even ifis odd, and can be odd or even ifis even.

$(A + B + 1)^{A+B+1}$ is odd ifis odd, and even ifis even.

$(A + B + 1)^{A+B+1}$ is even ifis even, and can be odd or even ifis odd.

$(A + B + 1)^{A+B+1}$ is odd ifis odd, and can be odd or even ifis even.

Correct answer:

$(A + B + 1)^{A+B+1}$ is odd ifis odd, and even ifis even.

Explanation:

Ifis odd, then A + B +1 , the sum of three odd integers, is odd; an odd number taken to any positive integer power is odd.

Ifis even, then A + B +1 ,两个奇数和偶数的和,is even; an even number taken to any positive integer power is even.

Therefore, $(A + B + 1)^{A+B+1}$ always assumes the same odd/even parity as.

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SSAT Upper Level Math : Properties of Exponents

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #121 :How To Find The Properties Of An Exponent

Example Question #122 :How To Find The Properties Of An Exponent

Example Question #123 :How To Find The Properties Of An Exponent

Example Question #124 :How To Find The Properties Of An Exponent

Example Question #125 :How To Find The Properties Of An Exponent

Example Question #181 :Algebra

Example Question #182 :Algebra

Example Question #183 :Algebra

Example Question #184 :Algebra

Example Question #121 :Properties Of Exponents

All SSAT Upper Level Math Resources

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