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SAT II Math II : SAT Subject Test in Math II

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 #11 : Exponents And Logarithms

Simplify:

$\frac{(a^3b^2c^7)^{-3}}{(a^7b^4c^{-2})^2}$

Possible Answers:

$\frac{b^4}{c^17}$

$\frac{a^2}{b^2c^{17}}$

$\frac{1}{a^{23}b^{14}c^{17}}$

$\frac{b^5}{a^{10}c^{21}}$

Correct answer:

$\frac{1}{a^{23}b^{14}c^{17}}$

Explanation:

Start by distributing the exponent in both the numerator and the denominator. Recall that when an exponent is raised to the exponent, you will need to multiply the two numbers together.

$\frac{(a^3b^2c^7)^{-3}}{(a^7b^4c^{-2})^2}=\frac{a^{-9}b^{-6}c^{-21}}{a^{14}b^8c^{-4}}$

Next, recall that when you have numbers with different exponents, but the same base, subtract the exponent found in the denominator from the exponent in the numerator.

$\frac{a^{-9}b^{-6}c^{-21}}{a^{14}b^8c^{-4}}=a^{-23}b^{-14}c^{-17}$

Recall that you can flip the fraction to make the exponents positive.

$a^{-23}b^{-14}c^{-17}=\frac{1}{a^{23}b^{14}c^{17}}$

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Example Question #17 : Exponents And Logarithms

Solve $(\frac{1}{36})^{x+3}=\frac{1}{216}$ .

Possible Answers:

$-\frac{5}{7}$

$\frac{5}{4}$

$\frac{1}{3}$

$-\frac{3}{2}$

Correct answer:

$-\frac{3}{2}$

Explanation:

The first thing we need to do is find a common base. This can be tricky to do, but guessing and checking a little shows that:

$(\frac{1}{216})=(\frac{1}{36})^{\frac{3}{2}}$

Plugging that back in to the original equation:

$(\frac{1}{36})^{x+3}=(\frac{1}{36})^{\frac{3}{2}}$

Now that our bases are the same, we can cancel them:

$x+3=\frac{3}{2}$

From here, it's much easier to solve using simple algebra:

$x=-\frac{3}{2}$

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Example Question #11 : Exponents And Logarithms

Solve 5 + log_2 (5x) = 9 .

Possible Answers:

$\frac{3}{e}$

$\frac{7}{4}$

$\frac{-3}{2}$

$\frac{3}{8}$

$\frac{16}{5}$

Correct answer:

$\frac{16}{5}$

Explanation:

First, we gather all the constants on one side of the equation:

log_2 (5x) = 4

Next, we rewrite the equation in exponential form:

2^4=5x

Now we can simplify the exponent:

16=5x

And finally, divide:

$\frac{16}{5}=x$

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Example Question #19 : Exponents And Logarithms

Solve log_9 10 - log_9 5x = 1 .

Possible Answers:

$\frac{3}{e}$

$\frac{4}{7}$

$\frac{5}{12}$

$\frac{2}{9}$

$\frac{3}{7}$

Correct answer:

$\frac{2}{9}$

Explanation:

Start by combining log terms. Remember, if you subtract logs, you divide the terms inside them:

$log_9(\frac{10}{5x})=1$

Now we can rewrite the equation in exponential form:

$9^1=\frac{10}{5x}$

Finally, we need to get the variable in the numerator, and then alone:

45x=10

$x=\frac{10}{45}=\frac{2}{9}$

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Example Question #51 : Sat Subject Test In Math Ii

Solve log_3(4x - 10) - log_3 8 = log_3 40 .

Possible Answers:

$\frac{17}{2}$

$\frac{165}{2}$

Correct answer:

$\frac{165}{2}$

Explanation:

First, we combine log terms by subtracting them. Remember, when you subtract logs, you divide the terms inside them:

$log_3(\frac{4x-10}{8})=log_340$

Now, because the bases of the logs match on either side of the equation, we can cancel them out:

$\frac{4x-10}{8}=40$

From here, we use simple algebra to solve:

4x-10=320

4x=330

$x=\frac{165}{2}$

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Example Question #21 : Mathematical Relationships

Solve ln x - ln 4 = 5

Possible Answers:

4+e^5

1024e

1024+e^5

4e^5

1024e^5

Correct answer:

4e^5

Explanation:

First, subtract the natural log terms:

$ln(\frac{x}{4})=5$

Now rewrite the equation in exponential form:

$e^5=\frac{x}{4}$

Finally, isolate the variable:

4e^5=x

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Example Question #21 : Mathematical Relationships

Solve $log_{18} (2 - x) = log_{18} (-2x - 2)$

Possible Answers:

Correct answer:

Explanation:

We can start by canceling the logs, because they both have the same base:

2-x=-2x-2

Now we can collect constants on one side of the equation, and variables on the other:

x=-4

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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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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #11 : Exponents And Logarithms

Example Question #17 : Exponents And Logarithms

Example Question #11 : Exponents And Logarithms

Example Question #19 : Exponents And Logarithms

Example Question #51 : Sat Subject Test In Math Ii

Example Question #21 : Mathematical Relationships

Example Question #21 : Mathematical Relationships

Example Question #21 : Exponents And Logarithms

Example Question #31 : Mathematical Relationships

Example Question #31 : Mathematical Relationships

All SAT II Math II Resources

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