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SAT Math : Exponents

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Example Questions

Example Question #2 : How To Divide Complex Numbers

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(2+i) \ast i$ .

Possible Answers:

$2+ \frac{ 6 }{5}i$

$\frac{ 2 }{3}+2i$

$\frac{ 6 }{5} + \frac{ 2 }{5}i$

$\frac{ 2 }{5} + \frac{ 6 }{5}i$

$2+ \frac{ 2 }{3}i$

Correct answer:

$\frac{ 6 }{5} + \frac{ 2 }{5}i$

Explanation:

$a \ast b = \frac{a + b}{a}$ , so

$(2+i) \ast i = \frac{(2+ i )+ i}{2+i} = \frac{2 + 2i}{2+i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 2 - i :

$(2+i) \ast i$

$= \frac{2 + 2i}{2+i}$

$= \frac{\left (2 + 2i \right )\left (2-i \right )}{\left (2+i \right )\left (2-i \right )}$

$= \frac{ 2 \cdot 2 - 2 \cdot i + 2i \cdot 2 - 2i \cdot i}{2 ^{2}+1^{2}}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot i^{2}}{4+1}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot (-1)}{5}$

$= \frac{ 4 - 2 i + 4i +2}{5}$

$= \frac{ 6 + 2 i }{5}$

$= \frac{ 6 }{5} + \frac{ 2 }{5}i$

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Example Question #602 : Algebra

Define an operation $\Phi$ such that, for any complex number ,

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

Possible Answers:

$\frac{ 2}{5} + \frac{1}{5} i$

$\frac{ 1}{3} + \frac{2}{3} i$

$\frac{ 2}{5} - \frac{1}{5} i$

$\frac{ 2}{3} + \frac{1}{3} i$

$\frac{ 1}{5} - \frac{2}{5} i$

Correct answer:

$\frac{ 2}{5} + \frac{1}{5} i$

Explanation:

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

N (1+ 2i) = i , and

$N=\frac{ i}{ 1+ 2i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 1 -2i :

$N=\frac{ i}{ 1+ 2i}$

$=\frac{ i( 1- 2i)}{( 1+ 2i)( 1- 2i)}$

$=\frac{ i \cdot 1- i \cdot 2i}{1 ^{2}+2^{2}}$

$=\frac{ i - 2i^{2}}{1+4}$

$=\frac{ i - 2(-1)}{5}$

$=\frac{ i +2}{5}$

$=\frac{ 2}{5} + \frac{1}{5} i$

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Example Question #3 : How To Divide Complex Numbers

Define an operation $\blacklozenge$ as follows:

For any two complex numbers and ,

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$

Evaluate $(1+ i) \blacklozenge (1+ i)$ .

Possible Answers:

$\frac{ 3 }{2} - \frac{1}{2} i$

$\frac{ 1 }{2} + \frac{3}{2} i$

$\frac{ 3 }{2} + \frac{1}{2} i$

$\frac{ 1 }{2} -\frac{3}{2} i$

Correct answer:

$\frac{ 3 }{2} + \frac{1}{2} i$

Explanation:

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$ , so

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

We can simplify each expression separately by rationalizing the denominators.

$\frac{1+ i}{2i}$

$= \frac{\left (1+ i \right )\cdot i}{2i \cdot i}$

$= \frac{1 \cdot i+ i \cdot i}{2 \cdot i ^{2}}$

$= \frac{ i+(-1)}{2 (-1)}$

$= \frac{-1+i}{-2}$

$= \frac{ 1 }{2} - \frac{1}{2} i$

$\frac{2i}{1+ i}$

$= \frac{2i(1- i)}{(1+ i)(1- i)}$

$= \frac{2i \cdot 1- 2i \cdot i}{1^{2}+1^{2}}$

$= \frac{2i - 2i ^{2}}{1+1}$

$= \frac{2i - 2(-1)}{2}$

$= \frac{2i +2}{2}$

= 1 + i

Therefore,

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

$=\left ( \frac{ 1 }{2} - \frac{1}{2} i \right )+ \left (1 + i \right )$

$=\left ( \frac{ 1 }{2}+1 \right )+ \left ( i- \frac{1}{2} i \right )$

$= \frac{ 3 }{2} + \frac{1}{2} i$

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Example Question #11 : Complex Numbers

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(3+2i) \ast 2$

Possible Answers:

$\frac{11 }{13} + \frac{ 16}{13}i$

$\frac{19 }{13} - \frac{ 4}{13}i$

$\frac{19}{13} + \frac{ 16}{13}i$

$\frac{11 }{13} - \frac{ 4}{13}i$

$\frac{19}{13} - \frac{ 16}{13}i$

Correct answer:

$\frac{19 }{13} - \frac{ 4}{13}i$

Explanation:

$a \ast b = \frac{a + b}{a}$ , so

$(3+2i) \ast 2 = \frac{(3+2i) + 2}{3+2i } = \frac{5+2i}{3+2i }$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 3 -2i :

$(3+2i) \ast 2 = \frac{5+2i}{3+2i }$

$= \frac{\left (5+2i \right )\left (3-2i \right )}{\left (3+2i \right )\left (3-2i \right )}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2i \cdot 2i }{3^{2}+2^{2}}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2 \cdot 2 \cdot i^{2} }{9+4}$

$= \frac{15 - 10i + 6i -4 \cdot (-1) }{9+4}$

$= \frac{15 - 10i + 6i +4}{13}$

$= \frac{19 -4i}{13}$

$= \frac{19 }{13} - \frac{ 4}{13}i$

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Example Question #2381 : Sat Mathematics

Define an operation $\Theta$ such that for any complex number ,

$\Theta a = 2a i + i$

If $\Theta N = 1,000$ , evaluate .

Possible Answers:

$500 + \frac{1}{2}i$

$\frac{ 1 }{2}+ 500i$

$- \frac{ 1 }{2}- 500i$

$- \frac{ 1 }{2}+ 500i$

$500 - \frac{1}{2}i$

Correct answer:

$- \frac{ 1 }{2}- 500i$

Explanation:

First substitute our variable N in where ever there is an a.

Thus, $\Theta a = 2a i + i$ , becomes $\Theta N = 2N i + i$ .

Since $\Theta N = 1,000$ , substitute:

2N i + i = 1,000

In order to solve for the variable we will need to isolate the variable on one side with all other constants on the other side. To do this, apply the oppisite operation to the function.

First subtract i from both sides.

2N i= 1,000 - i

Now divide by 2i on both sides.

$\frac{2N i}{2i}= \frac{1,000 - i}{2i}$

From here multiply the numerator and denominator by i to further solve.

$N= \frac{1,000 - i}{2i}$

$= \frac{(1,000 - i) \cdot i }{2i \cdot i}$

$= \frac{1,000\cdot i - i\cdot i }{2 \cdot i^{2}}$

Recall that i^2=-1 by definition. Therefore,

$= \frac{1,000 i - (-1) }{2 \cdot (-1)}$

$= \frac{1,000 i +1 }{-2}$

$= \frac{ 1 }{-2}+ \frac{1,000 i }{-2}$

$=- \frac{ 1 }{2}- 500i$ .

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Example Question #21 : Squaring / Square Roots / Radicals

Let 4x^2 = y^3 . What is the following equivalent to, in terms of :

$\frac{x}{\sqrt{y}}$

Possible Answers:

$\frac{1}{2}\sqrt{y}$

$\pm\frac{1}{8}y^{\frac{3}{2}}$

$\frac{1}{4}y^2$

$\pm\frac{1}{2}y$

Correct answer:

$\pm\frac{1}{2}y$

Explanation:

Solve for x first in terms of y, and plug back into the equation.

$\\4x^2 = y^3 \\ \\x^2 = \frac{1}{4} y^3 \\ \\x = \pm \frac{1}{2} y\sqrt{y}$

Then go back to the equation you are solving for:

$\frac{x}{\sqrt{y}}$ substitute in $x=\pm\frac{1}{2}y\sqrt{y}$

$\\ \frac{x}{\sqrt{y}}=\frac{\pm\frac{1}{2}y\sqrt{y}}{\sqrt{y}} \\ \\\frac{x}{\sqrt{y}}=\pm\frac{1}{2}y$

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Example Question #22 : Squaring / Square Roots / Radicals

Simplify the expression by rationalizing the denominator, and write the result in standard form: $\frac{5+i}{7-i}$

Possible Answers:

$\frac{17}{25} + \frac{ 6}{25}i$

$\frac{17}{24} + \frac{ 1}{4}i$

$\frac{18}{25} + \frac{ 6}{25}i$

$\frac{3}{4} + \frac{ 1}{4}i$

$\left (\textup{Note: } i=\sqrt{-1}\right\ )$

Correct answer:

$\frac{17}{25} + \frac{ 6}{25}i$

Explanation:

Multiply both numerator and denominator by the complex conjugate of the denominator, which is 7 + i :

$\frac{5+i}{7-i}$

$= \frac{(5+i)(7+i)}{(7-i)(7+i)}$

$= \frac{5 \cdot 7 + 5 \cdot i + 7 \cdot i + i \cdot i }{(7^{2}+1 ^{2}) }$

$= \frac{35 +5i + 7 i +(-1) }{(49+1) }$

$= \frac{34 +12i }{50 }$

$= \frac{34 }{50 } + \frac{ 12i }{50 }$

$= \frac{17}{25} + \frac{ 6}{25}i$

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Example Question #2381 : Sat Mathematics

Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one.

Possible Answers:

24 + 7i

7 + i

12 - 12i

Correct answer: 24 + 7i

Explanation:

Distribute (3 + 4i)(4 - 3i)

3(4) + 3(-3i) + 4i(4) + 4i(-3i)

12 - 9i + 16i -12i²

12 + 7i - 12(-1)

12 + 7i + 12

24 + 7i

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Example Question #2382 : Sat Mathematics

$x^{4}-1$ has 4 roots, including the complex numbers. Take the product of $\left ( 5+2i \right )$ with each of these roots. Take the sum of these 4 results. Which of the following is equal to this sum?

Possible Answers:

The correct answer is not listed.

3+7i

4-10i

10i

Correct answer:

Explanation:

$x^{4}-1=\left ( x^{2} -1\right )\left ( x^{2}+1 \right )=\left ( x+1 \right )\left ( x-1 \right )\left ( x-i \right )\left ( x+i \right )$

This gives us roots of

$1,\ -1,\ i,\ -i$

The product of $\left ( 5+2i \right )$ with each of these gives us:

$(5+2i)\cdot i = 5i + 2i^2 = 5i +2(-1) = 5i-2 = -2 + 5i$

$(5+2i)\cdot (-i) = -5i -2i^2 = -5i -2(-1) = -5i +2 = 2-5i$

$(5+2i)\cdot 1 = 5+2i$

$(5+2i)\cdot (-1) = -5 -2i$

The sum of these 4 is:

[(-2+5i) + (2-5i)] + [(5+2i) + (-5-2i)] =

=[(-2)+2 +5i - 5i] + [5-5 + 2i - 2i] = [0] + [0] = 0

What we notice is that each of the roots has a negative. It thus makes sense that they will all cancel out. Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property:

$1(5+2i) + (-1)\cdot(5+2i)+ i(5+2i)+(-i) \cdot(5+2i) = (5+2i)\cdot[1+(-1)+i+(-i)]$

$= (5+2i)\cdot[0] = 0$

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Example Question #2386 : Sat Mathematics

Simplify:

$(3i)^{6}$

Possible Answers:

729 i

-729 i

None of the other responses gives the correct answer.

729

Correct answer:

Explanation:

Apply the Power of a Product Property:

$(3i)^{6} = 3 ^{6} \cdot i^{6} = 729 \cdot i^{6}$

A power of can be found by dividing the exponent by 4 and noting the remainder. 6 divided by 4 is equal to 1, with remainder 2, so

$i^{6} = i^{2} = -1$

Substituting,

$(3i)^{6} = 729 \cdot (-1) = -729$ .

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SAT Math : Exponents

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2 : How To Divide Complex Numbers

Example Question #602 : Algebra

Example Question #3 : How To Divide Complex Numbers

Example Question #11 : Complex Numbers

Example Question #2381 : Sat Mathematics

Example Question #21 : Squaring / Square Roots / Radicals

Example Question #22 : Squaring / Square Roots / Radicals

Example Question #2381 : Sat Mathematics

Example Question #2382 : Sat Mathematics

Example Question #2386 : Sat Mathematics

All SAT Math Resources

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