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SAT II Math I : Mathematical Relationships

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Example Questions

Example Question #11 : Exponents And Logarithms

Evaluate: $xe^{ln(4-2x)}$

Possible Answers:

(4x-2)^x

-2x^2-4x

-2x^2+4x

4x-2

2x-4

Correct answer:

-2x^2+4x

Explanation:

An exponential base raised to the natural log will eliminate, leaving only the terms of the power. This is a log rule that can be used to simplify the expression.

$xe^{ln(4-2x)} = x(4-2x)$

Distribute the x variable through the binomial.

4x-2x^2

The answer is: -2x^2+4x

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Example Question #15 : Exponents And Logarithms

$\textup{Solve for }x:$

$2 ^{2x} +4 = 2^{x+2}$

Possible Answers:

$\textup{The equation has no solution}$

x = 1

x = - 1

x = 0

x = 2

Correct answer:

x = 1

Explanation:

By the Power of a Power and Product of Power Rules, we can rewrite this equation as

$2 ^{2x} +4 = 2^{x+2}$

$(2 ^{ x }) ^{2} +4 = 2^{2} \cdot 2^{x}$

$(2 ^{ x }) ^{2} +4 = 4 \cdot 2^{x}$

Substitute for $2^{x}$ ; the resulting equation is the quadratic equation

$t^{2} +4 = 4t$ ,

which can be written in standard form by subtracting from both sides:

$t^{2} +4 - 4t = 4t - 4t$

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

The quadratic trinomial fits the perfect square trinomial pattern:

$t^{2}- 2 \cdot t \cdot 2 + 2^{2} = 0$

$(t-2)^{2} = 0$

By the square root principle,

t - 2= 0

t = 2

Substituting $2^{x}$ for :

$2^{x} = 2$

$2^{x} = 2 ^{1}$

x = 1

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Example Question #16 : Exponents And Logarithms

Solve for (round to the nearest hundredth):

$3 ^{x+ 4} = 2^{x+ 9}$

Possible Answers:

0.45

1.03

0.18

26.22

5.93

Correct answer:

0.45

Explanation:

Take the natural logarithm of both sides:

$3 ^{x+ 4} = 2^{x+ 9}$

$\ln 3 ^{x+ 4} = \ln 2^{x+ 9}$

By Logarithm of a Power Rule, the above becomes

$(x+ 4) \ln 3 = (x+ 9) \ln 2$

After distributing, solve for :

$x \ln 3 + 4 \ln 3 = x \ln 2+ 9 \ln 2$

$x \ln 3 + 4 \ln 3 - x \ln 2 - 4 \ln 3 = x \ln 2+ 9 \ln 2 - x \ln 2 - 4 \ln 3$

$x \ln 3 - x \ln 2 = 9 \ln 2 - 4 \ln 3$

Factor out the left side, then divide:

$x( \ln 3 - \ln 2) = 9 \ln 2 - 4 \ln 3$

$\frac{x( \ln 3 - \ln 2) }{\ln 3 - \ln 2}=\frac{ 9 \ln 2 - 4 \ln 3}{\ln 3 - \ln 2}$

$x=\frac{ 9 \ln 2 - 4 \ln 3}{\ln 3 - \ln 2}$

Substituting the values of the logarithms:

$x \approx \frac{ 9 (0.6931) - 4 (1.0986)}{1.0986 - 0.6931}$

$x\approx \frac{ 1.8439}{0.4055}$

$x \approx 0.45472$

This rounds to 0.45.

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Example Question #62 : Sat Subject Test In Math I

Solve for :

$2^{x} = 8 ^{2-x}$

Possible Answers:

No solution

x = 2

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

x = 3

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

Correct answer:

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

Explanation:

$8 = 2^{3}$ , so the equation

$2^{x} = 8 ^{2-x}$

can be rewritten as:

$2^{x} =( 2 ^{3} ) ^{2-x}$

By the Power of a Power rule:

$2^{x} = 2 ^{3 (2-x)}$

It follows that

x = 3 (2-x )

Solving for :

$x = 3 \cdot 2- 3 \cdot x$

x = 6 - 3x

x + 3x = 6 - 3x + 3x

4x = 6

$4x \div 4 = 6 \div 4$

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

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Example Question #1 : Absolute Value

Define an operation $\vee$ on the set of real numbers as follows:

For any two real numbers a, b

$a \vee b =| | a + 2b | + |2a + b| |$

Evaluate the expression

$4 \vee\left ( -4 \right )$

Possible Answers:

Correct answer:

Explanation:

Substitute a= 4, b = -4 in the expression:

$a \vee b =| | a + 2b | + |2a + b| |$

$4 \vee\left ( -4 \right ) =| | 4 + 2 (-4) | + |2 (4) + (-4)| |$

$4 \vee\left ( -4 \right ) =| | 4 + (-8) | + |8+ (-4)| |$

$4 \vee\left ( -4 \right ) =| | -4 | + |4| |$

$4 \vee\left ( -4 \right ) =|4+4|$

$4 \vee\left ( -4 \right ) =|8|$

$4 \vee\left ( -4 \right ) =8$

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Example Question #1 : Absolute Value

Simplify the following expression:

$\mid -5x\mid -4+6x-\mid -3\mid$

Possible Answers:

x-7

11x-7

11x-1

x-1

Correct answer:

11x-7

Explanation:

$\mid -5x\mid -4+6x-\mid -3\mid$

To simplify, we must first simplify the absolute values.

5x -4+6x-3

Now, combine like terms:

11x-7

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Example Question #2 : Absolute Value

Solve for .

$\left | 2x+3\right |=7$

Possible Answers:

x=10,3

x=-5, 2

x=5, 2

x=5,-2

Correct answer:

x=-5, 2

Explanation:

To solve for x we need to make two separate equations. Since it has absolute value bars around it we know that the inside can equal either 7 or -7 before the asolute value is applied.

$\left | 2x+3\right |=7$

2x +3=7 2x+3=-7

2x=4 2x=-10

x=2 x=-5

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Example Question #4 : Absolute Value

The absolute value of a negative can be positive or negative. True or false?

Possible Answers:

True

False

Correct answer:

False

Explanation:

The absolute value of a number is the points away from zero on a number line.

Since this is a countable value, you cannot count a negative number.

This makes all absolute values positive and also make the statement above false.

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Example Question #2 : Absolute Value

Consider the quadratic equation

$x^{2} +14 = 9x$

Which of the following absolute value equations has the same solution set?

Possible Answers:

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

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

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

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

None of the other choices gives the correct response.

Correct answer:

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

Explanation:

Rewrite the quadratic equation in standard form by subtracting from both sides:

$x^{2} +14 = 9x$

$x^{2} +14 - 9x = 9x - 9x$

$x^{2}- 9x +14 = 0$

Factor this as

$(x+ \square )(x+ \square ) = 0$

where the squares represent two integers with sum and product 14. Through some trial and error, we find that and work:

(x-2)(x-7) = 0

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

If x- 2= 0 then x= 2 ;

if x- 7= 0 then x= 7 .

The given equation has solution set $\left \{ 2, 7\right \}$ , so we are looking for an absolute value equation with this set as well.

This equation can take the form

|x-a| = b

This can be rewritten as the compound equation

x-a = -b x-a = b

Adding to both sides of each equation, the solution set is

x =a -b and x =a + b

Setting these numbers equal in value to the desired solutions, we get the linear system

a -b = 2

a+ b = 7

Adding and solving for :

a -b = 2

$\underline{a+ b = 7}$

= 9

$2a \div 2 = 9 \div 2$

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

Backsolving to find :

a+ b = 7

$4 \frac{1}{2} + b = 7$

$4 \frac{1}{2} + b - 4 \frac{1}{2} = 7 - 4 \frac{1}{2}$

$b =2 \frac{1}{2}$

The desired absolute value equation is $\left |x- 4 \frac{1}{2} \right |= 2 \frac{1}{2}$ .

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Example Question #2 : Absolute Value

What is the value of: $\left | -(2^{11}) \right |$ ?

Possible Answers:

1024

4096

2048

Correct answer:

2048

Explanation:

Step 1: Evaluate $2^{11}$ ...

$2^{11}=2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2=2048$

Step 2: Apply the minus sign inside the absolute value to the answer in Step 1...

$2048\rightarrow-2048$

Step 3: Define absolute value...

The absolute value of any value $\pm a$ is always positive, unless there is an extra negation outside (sometimes)..

Step 4: Evaluate...

$\left | (-2048)\right |=2048$

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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 #11 : Exponents And Logarithms

Example Question #15 : Exponents And Logarithms

Example Question #16 : Exponents And Logarithms

Example Question #62 : Sat Subject Test In Math I

Example Question #1 : Absolute Value

Example Question #1 : Absolute Value

Example Question #2 : Absolute Value

Example Question #4 : Absolute Value

Example Question #2 : Absolute Value

Example Question #2 : Absolute Value

All SAT II Math I Resources

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