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SAT II Math II : Mathematical Relationships

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

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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #21 : Exponents And Logarithms

Solve -log_7(x - 7) + 1 = -1 .

Possible Answers:

Correct answer:

Explanation:

We can start by gathering all the constants to one side of the equation:

-log_7(x - 7) = -2

Next, we can multiply by to change the signs:

log_7(x - 7) = 2

Now we can rewrite the equation in exponential form:

7^2=x-7

And finally, we can solve algebraically:

49=x-7

56=x

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Example Question #31 : Mathematical Relationships

Solve for :

$81^{x-3}=27^{2x}$

Possible Answers:

x=-5

x=-4

x=2

x=-6

Correct answer:

x=-6

Explanation:

In order to solve this problem, rewrite both sides of the equation in terms of raising to an exponent.

Since, 81=3^4 , we can write the following:

$81^{x-3}=(3^4)^{x-3}=3^{4x-12}$

Since 27=3^3 , we can write the following:

$27^{2x}=(3^3)^{2x}=3^{6x}$

Now, we can solve for with the following equation:

6x=4x-12

2x=-12

x=-6

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Example Question #31 : Mathematical Relationships

Solve

$-9e^{x + 1} = 47$

Possible Answers:

$\frac{3}{e}$

$ln(\frac{3}{4})$

$\frac{e}{4}$

No solutions

Correct answer:

No solutions

Explanation:

The first thing we need to do is find a common base. However, because one of the bases has an in it (an irrational number), and the other does not, it's going to be impossible to find a common base. Therefore, the question has no solution.

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Example Question #21 : Exponents And Logarithms

Solve $log_{15} (48 + x) = log_{15} (x^2 + 3x)$

Possible Answers:

6, 8

-8, 6

-8, -6

-6, 8

No solutions

Correct answer:

-8, 6

Explanation:

First, we can simplify by canceling the logs, because their bases are the same:

48+x=x^2+3x

Now we collect all the terms to one side of the equation:

x^2+2x-48=0

Factoring the expression gives:

(x-6)(x+8)=0

So our answers are:

x=-8, 6

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Example Question #21 : Exponents And Logarithms

Solve 25^x = 625^x .

Possible Answers:

x^2

No solutions

Correct answer:

Explanation:

Here, we can see that changing base isn't going to help. However, if we remember that and number raised to the th power equals , our solution becomes very easy.

$25^{(0)}=625^{(0)}$

1=1

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Example Question #21 : Exponents And Logarithms

To the nearest hundredth, solve for : $7^{2x} = 5 ^{x}$ .

Possible Answers:

2.28

None of these

0.55

1.27

0.99

Correct answer:

None of these

Explanation:

Take the natural logarithm of both sides:

$7^{2x} = 5 ^{x}$

$\ln 7^{2x} = \ln 5 ^{x}$

By the Logarithm of a Power Rule the above becomes

$2x \ln 7 = x \ln 5$

Solve for :

$2x \ln 7 - x \ln 5 = x \ln 5 - x \ln 5$

$(2 \ln 7 - \ln 5) x = 0$

$\frac{(2 \ln 7 - \ln 5) x }{2 \ln 7 - \ln 5}= \frac{0 }{2 \ln 7 - \ln 5}$

x = 0 .

This is not among the choices given.

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Example Question #1 : Absolute Value

Define $g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$ .

Evaluate g(25) .

Possible Answers:

1,630

940,900

380

1,620

370

Correct answer:

370

Explanation:

$g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$

$g(25)= \left | \left | 1,000- \sqrt{25}\right | - 25^{2} \right |$

$= \left | \left | 1,000-5\right | - 625 \right |$

$= \left | \left | 995 \right | - 625 \right |$

$= \left | 995 - 625 \right |$

$= \left | 370 \right |$

=370

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Example Question #1 : Absolute Value

Define $f(x) = \left | 1 - x\right |$ .

Order from least to greatest: $f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Possible Answers:

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

$f\left ( \frac{13}{14} \right ), f\left ( \frac{9}{14} \right ), f\left ( \frac{17}{14} \right )$

$f\left ( \frac{17}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{9}{14} \right )$

$f\left ( \frac{9}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{13}{14} \right )$

$f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Correct answer:

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

Explanation:

$f(x) = \left | 1 - x\right |$ , or, equivalently,

$f(x) = \left | \frac{14}{14} - x\right |$

$f\left ( \frac{9}{14} \right ) = \left | \frac{14}{14} - \frac{9}{14}\right | = \left | \frac{5}{14} \right |=\frac{5}{14}$

$f\left ( \frac{13}{14} \right ) = \left | \frac{14}{14} - \frac{13}{14}\right |= \left | \frac{1}{14} \right |=\frac{1}{14}$

$f\left ( \frac{17}{14} \right ) = \left | \frac{14}{14} - \frac{17}{14} \right |= \left | - \frac{3}{14} \right |=\frac{3}{14}$

From least to greatest, the values are

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

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Example Question #3 : Absolute Value

Define an operation $\Psi$ as follows:

For all real numbers a , b ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

Evaluate $2\frac{1}{2} \Psi \left (- \frac{1}{4} \right )$ .

Possible Answers:

$1\frac{1}{5}$

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

Undefined.

Correct answer:

Explanation:

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

$2\frac{1}{2} \Psi \left (- \frac{1}{4} \right ) = \frac{\left |1- 2\frac{1}{2} \right |}{1-\left |-\frac{1}{4}\right |}$

$= \frac{\left |-1\frac{1}{2} \right |}{1- \frac{1}{4}}$

$= \frac{ 1\frac{1}{2} }{ \frac{3}{4}}$

$= \frac{ \frac{3}{2} }{ \frac{3}{4}}$

$= \frac{3}{2} \div \frac{3}{4}$

$= \frac{3}{2} \cdot \frac{4}{3}$

$= \frac{12}{6}$

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Example Question #61 : Sat Subject Test In Math Ii

Define an operation $\Psi$ as follows:

For all real numbers a , b ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

If $\frac{1}{2}\Psi N = 1$ , which is a possible value of ?

Possible Answers:

N =- 3

N =- 2

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

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

N =- 1

Correct answer:

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

Explanation:

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$ , so

$\frac{1}{2}\Psi N = 1$

can be rewritten as

$\frac{\left |1- \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{\left | \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{ \frac{1}{2} }{1-\left |N \right |} = 1$

$\frac{1}{2} =1-\left |N \right |$

$-\frac{1}{2} =-\left |N \right |$

$\frac{1}{2} = \left |N \right |$

Therefore, either $N = \frac{1}{2}$ or $N =- \frac{1}{2}$ . The correct choice is $N =- \frac{1}{2}$ .

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SAT II Math II : Mathematical Relationships

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

All SAT II Math II Resources

Example Questions

Example Question #21 : Exponents And Logarithms

Example Question #31 : Mathematical Relationships

Example Question #31 : Mathematical Relationships

Example Question #21 : Exponents And Logarithms

Example Question #21 : Exponents And Logarithms

Example Question #21 : Exponents And Logarithms

Example Question #1 : Absolute Value

Example Question #1 : Absolute Value

Example Question #3 : Absolute Value

Example Question #61 : Sat Subject Test In Math Ii

All SAT II Math II Resources

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