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SAT II Math I : Mathematical Relationships

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

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Example Question #51 : Sat Subject Test In Math I

Simplify $(x^{-4})^7$

Possible Answers:

$x^{-47}$

$x^{\frac{7}{4}}$

$x^{-11}$

x^3

$x^{-28}$

Correct answer:

$x^{-28}$

Explanation:

When an exponent is raised by another exponent, we just multiply the powers.

$(x^{-4})^7=x^{-4*7}=x^{-28}$

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Example Question #1 : Exponents And Logarithms

Simplify:

3^9+3^9+3^9

Possible Answers:

$27^{9}$

$3^{10}$

$9^{27}$

9^9

$3^{27}$

Correct answer:

$3^{10}$

Explanation:

When adding exponents, we don't add the exponents or multiply out the bases. Our goal is to see if we can factor anything. We do see three 3^9 . Let's factor.

$3^9+3^9+3^9=3^9(1+1+1)=3^9(3)=3^{10}$ Remember when multiplying exponents, we just add the powers.

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Example Question #1 : Exponents And Logarithms

Solve and simplify.

$\sqrt[3]{125}$

Possible Answers:

$5\sqrt{5}$

$5\sqrt[3]{5}$

Correct answer:

Explanation:

$\sqrt[3]{125}$ Another way to write this is x^3=125 . The only number that makes 125 is 5^3 .

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Example Question #7 : Exponents And Logarithms

Simplify:

$\log_3729$

Possible Answers:

$3^{729}$

3^9

243

Correct answer:

Explanation:

$\log_3729$ is the same as 3^x=729 . Let's factor out 729 . It's the same as 3^6 . Therefore x=6 which is the answer to our question.

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Example Question #1 : Exponents And Logarithms

Simplify:

$\log50-\log5$

Possible Answers:

$5\log50$

$\log250$

$\log45$

Correct answer:

Explanation:

When dealing with subtraction in regards to logarithms, it's the same as dividing the numbers.

$\log50-\log5=\frac{\log50}{\log5}=\log \frac{50}{5}=\log10=1$

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Example Question #2 : Exponents And Logarithms

Simplify:

$\log_618+\log_672$

Possible Answers:

$\log_690$

$6\log90$

$3\log_618$

Correct answer:

Explanation:

When dealing with addition in regards to logarithms, it's the same as multiplying the numbers.

$\log_618+\log_672=\log_6(18*72)=\log_61296=4$

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Example Question #10 : Exponents And Logarithms

Solve: $(x^{2})^{5}$ when x=2 .

Possible Answers:

1024

100

1000

Correct answer:

1024

Explanation:

Power rule says when an exponent is raised to another exponent, you must multiply the exponents.

So 2*5=10 and our expression is now $x^{10}$ .

Plug in the given value to get

$2^{10}=1024$ .

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Example Question #61 : Sat Subject Test In Math I

Give the set of real solutions to the equation

$3 ^{2x+1} -28 \cdot 3^{x}+9= 0$

(round to the nearest hundredth, if applicable)

Possible Answers:

$\left \{ -1, 2 \right \}$

$\left \{ 2 \right \}$

$\left \{ 1, 2 \right \}$

$\left \{ -1 \right \}$

The equation has no real solutions.

Correct answer:

$\left \{ -1, 2 \right \}$

Explanation:

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:

$3 ^{2x+1} -28 \cdot 3^{x}+9= 0$

$3 ^{2x } \cdot 3^{1 } -28 \cdot 3^{x}+9= 0$

$3 \cdot 3 ^{2x } -28 \cdot 3^{x}+9= 0$

$3 \cdot (3 ^{x } )^{2 } -28 \cdot 3^{x}+9= 0$

Substitute for $3^{x}$ ; the equation becomes

$3t^{2 } -28t+9= 0$

which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is $3 \cdot 9 = 27$ and whose sum is . By trial and error, we find the integers to be and , so the equation can be rewritten as follows:

$3t^{2 } -27t -t +9= 0$

Factoring by grouping:

$(3t^{2 } -27t )-( t -9)= 0$

3t (t -9 )-1 ( t -9)= 0

(3t -1 )( t -9)= 0

By the Zero Product Rule, one of these two factors must be equal to 0.

If t - 9 = 0 , then t = 9 .

Since $3 ^{2} = 9$ , then substituting this as well as substituting $3^{x}$ back for , we get

$3^{x} = 3 ^{2}$ ,

and

x = 2

If 3t-1 = 0 , then

3t = 1

$t = \frac{1}{3}$

Since $3 ^{-1} = \frac{1}{3}$ , then substituting this as well as substituting $3^{x}$ back for , we get

$3^{x} = 3 ^{-1}$ , and

x= -1

The solution set is therefore $\left \{ -1, 2 \right \}$

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Example Question #11 : Exponents And Logarithms

Give the set of real solutions to the equation

$2 ^{2x+2} + 11 \cdot 2^{x} - 3 = 0$

(round to the nearest hundredth, if applicable)

Possible Answers:

$\left \{ -2, -0.58\right \}$

$\left \{ -0.58\right \}$

$\left \{ -2, 0.58\right \}$

$\left \{ 0.58\right \}$

$\left \{ -2 \right \}$

Correct answer:

$\left \{ -2 \right \}$

Explanation:

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:

$2 ^{2x+2} + 11 \cdot 2^{x} - 3 = 0$

$2 ^{2x } \cdot 2 ^{2}+ 11 \cdot 2^{x} - 3 = 0$

$4 \cdot( 2 ^{x } ) ^{2}+ 11 \cdot 2^{x} - 3 = 0$

Substitute for $2^{x}$ ; the equation becomes

$4t ^{2}+ 11t- 3 = 0$ ,

which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is $4 \cdot (-3) = -12$ and whose sum is 11. By trial and error, we find the integers to be 12 and , so the equation can be written as follows:

$4t ^{2} -t + 12t- 3 = 0$

Factoring by grouping:

$(4t ^{2} -t )+( 12t- 3 )= 0$

t (4t -1 )+ 3 (4t -1 )= 0

(t + 3 )(4t -1 )= 0

By the Zero Product Rule, one of these two factors must be equal to 0.

If t+3 , then t = -3 .

Substituting $2^{x}$ back for , we get

$2^{x}= -3$ .

This is impossible, since any power of a positive number must be positive.

If 4t -1 = 0 , then:

4t =1

$t = \frac{1}{4}$

Substituting $2^{x}$ back for , we get

$2^{x}= \frac{1}{4}$

Since $\frac{1}{4} = \frac{1}{2^{2}} = 2^{-2}$ ,

it holds that $2^{x}= 2^{-2}$ , and x = -2 , the only solution.

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Example Question #11 : Exponents And Logarithms

Simplify $log_{10}(\frac{10}{100})$

Possible Answers:

Correct answer:

Explanation:

One of the properties of log is that $\small log_{x}(\frac{n}{m})=log_{x}(n)-log_{x}(m)$

Applying that principle to this problem:

$log_{10}(\frac{10}{100})=log_{10}(10)-log_{10}(100)$

Simplifying the log base 10

$log_{10}(10)=log_{10}(10^1)=1$

$log_{10}(100)=log_{10}(10^2)=2$

Plug in the values to the first equation:

$log_{10}(\frac{10}{100})=log_{10}(10)-log_{10}(100)=1-2=-1$

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All SAT II Math I Resources

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SAT II Math I : Mathematical Relationships

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #51 : Sat Subject Test In Math I

Example Question #1 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

Example Question #7 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

Example Question #2 : Exponents And Logarithms

Example Question #10 : Exponents And Logarithms

Example Question #61 : Sat Subject Test In Math I

Example Question #11 : Exponents And Logarithms

Example Question #11 : Exponents And Logarithms

All SAT II Math I Resources

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