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SSAT Upper Level Math : Properties of Exponents

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Example Questions

Example Question #562 : Grade 6

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

Possible Answers:

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

23^6

$6\times23$

$6^{23}$

$23\times6$

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 itself times, so that is our exponent.

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Example Question #171 : Algebra

$x^{5} = - 77$

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 #172 : Algebra

y > 0 and z> 0

$y^{2} = 12$

$z^{2} = 5$

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

Possible Answers:

$12\sqrt{15}$

$120 \sqrt{30 }$

$120 \sqrt{15}$

$12\sqrt{30}$

Correct answer:

$120 \sqrt{15}$

Explanation:

$y^{2} = 12$ and is positive, so

$y = \sqrt{12}$ .

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

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

$z^{2} = 5$ , and is 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 #173 : Algebra

$x^{2} = 75$

$y^{2} = 3$

and are both positive.

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

Possible Answers:

1,197

1,083

297

183

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 #174 : Algebra

$y ^{5} = -7$

$y^{20} = ?$

Possible Answers:

-2,401

2,401

Correct answer:

2,401

Explanation:

By the Power of a Power Principle,

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

Substituting for $y ^{5}$ , 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 #124 : Properties Of Exponents

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, since and are 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 #181 : 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, since and are 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 #182 : 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, since is 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

and are 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 odd if is odd, and even if is even.

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

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

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

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

Correct answer:

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

Explanation:

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

If is even, then A + B +1 , the sum of two odd integers and an even integer, 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 #562 : Grade 6

Example Question #171 : Algebra

Example Question #172 : Algebra

Example Question #173 : Algebra

Example Question #174 : Algebra

Example Question #124 : Properties Of Exponents

Example Question #181 : Algebra

Example Question #183 : Algebra

Example Question #182 : Algebra

Example Question #121 : Properties Of Exponents

All SSAT Upper Level Math Resources

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