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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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All SAT II Math II Resources

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

Example Questions

Example Question #1 : Exponents And Logarithms

To the nearest hundredth, solve for :

$2^{x+10}= 10^{x+2}$

Possible Answers:

$x \approx 3.85$

$x \approx 1.45$

$x \approx 2.31$

The equation has no solution.

$x\approx 7.17$

Correct answer:

$x \approx 1.45$

Explanation:

Take the common logarithm of both sides, then solve the resulting linear equation.

$2^{x+10}= 10^{x+2}$

$\log 2^{x+10}= \log 10^{x+2}$

$\left (x+10 \right )\log 2 = x+2$

$x\log 2 +10 \log 2 = x+2$

$x\log 2 +10 \log 2 - 2 - x \log 2= x+2 - 2 - x \log 2$

$-2 +10 \log 2 = x - x \log 2$

$x \left (1 - \log 2 \right )= -2 +10 \log 2$

$x =\frac{ -2 +10 \log 2}{1 - \log 2 }$

$x \approx \frac{ -2 +10 \cdot 0.3010}{1 - 0.3010 } \approx 1.45$

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Example Question #2 : Exponents And Logarithms

Solve for :

$100 ^{x}= 66$

Possible Answers:

$x = \log 33$

$x = \frac{2}{\log 66}$

$x = 2- \log 66$

$x = \frac{ \log 66}{2}$

$x = \log 64$

Correct answer:

$x = \frac{ \log 66}{2}$

Explanation:

Take the common logarithm of both sides:

$100 ^{x}= 66$

$\log 100 ^{x}= \log 66$

$x \log 100 = \log 66$

$x \cdot 2 = \log 66$

$x = \frac{ \log 66}{2}$

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Example Question #5 : Exponents And Logarithms

Solve for :

$3 + \log 4 + \log 6 = \log N$

Possible Answers:

N = 24,000

N = 1,000

N = 72

N = 10,000

N = 13,824

Correct answer:

N = 24,000

Explanation:

The base of the common logarithm is 10, so

$3 = \log 10^{3} = \log 1,000$

The sum of three logarithms is the logarithm of the product of the three powers, so:

$3 + \log 4 + \log 6$

$=\log 1,000 + \log 4 + \log 6$

$= \log (1,000 \times 4 \times 6) = \log 24,000$

Therefore, N = 24,000

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Example Question #1 : Exponents And Logarithms

To the nearest hundredth, solve for :

$5^{N} = 10^{4-N}$

Possible Answers:

$N \approx 0.37$

$N \approx 1.53$

$N \approx 1.30$

$N \approx 2.35$

$N \approx 13.29$

Correct answer:

$N \approx 2.35$

Explanation:

$5^{N} = 10^{4-N}$

$\log 5^{N} = \log 10^{4-N}$

$N \log 5 = 4 - N$

$N \log 5 + N = 4$

$N \left (1+ \log 5 \right ) = 4$

$N = \frac{4}{1+ \log 5 } \approx \frac{4}{1+ 0.6990 } \approx 2.35$

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Example Question #5 : Exponents And Logarithms

Solve for :

$\log 20 - 2 + \log 6 = \log N$

Possible Answers:

N = 60

$N = 1 \frac{1}{5}$

$N = \frac{60}{e}$

$N = \frac{120}{e^{2}}$

The equation has no solution

Correct answer:

$N = 1 \frac{1}{5}$

Explanation:

The base of the common logarithm is 10, so

$2 = \log 10^{2} = \log 1 00$

The sum of logarithms is the logarithm of the product of the three powers, and the difference of logarithms is the logarithm of the quotient of their powers. Therefore,

$\log 20 - 2 + \log 6$

$= \log 20 - \log 100 + \log 6$

$= \log \frac{20 \times 6 }{100} = \log \frac{120 }{100} =\log \frac{6}{5}$

$N = \frac{6}{5} = 1 \frac{1}{5}$

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Example Question #41 : Sat Subject Test In Math Ii

Give the set of real solutions to the equation

$2^{2x+ 4} - 2 ^{x+4} + 3 = 0$

(round to the nearest hundredth, if applicable)

Possible Answers:

$\left \{ -2 , 0.42 \right \}$

The equation has no solution.

$\left \{ -0.42, 0.42 \right \}$

$\left \{ -0.42, -2 \right \}$

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

Correct answer:

$\left \{ -0.42, -2 \right \}$

Explanation:

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first two terms strategically:

$2^{2( x+2)} - 2 ^{(x+2)+ 2 } + 3 = 0$

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

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

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

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

Factor this as

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

by finding two integers whose product is 3 and whose sum is . Through some trial and error we find -1, -3 , so we can write

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

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

If t- 1 = 0 , then t = 1 .

Substituting $2^{x+2 }$ back for :

$2^{x+2 } = 1$

$2^{x+2 } = 2^{0}$

x+2 = 0

x = -2 .

If t- 3 = 0 , then t = 3 .

Substituting $2^{x+2 }$ back for :

$2^{x+2 } = 3$

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

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

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

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

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

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

$x \approx \frac{1.0986 - 2 (0.6931 )}{ 0.6931 }$

$x \approx -0.42$

Both can be confirmed to be solutions by substitution.

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Example Question #41 : Sat Subject Test In Math Ii

Give the solution set:

$6^{2x+1} + 6 = 6 ^{x+2} + 6 ^{x}$

Possible Answers:

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

$\left \{ 6\right \}$

$\left \{ 1 \right \}$

The equation has no solution.

$\left \{ \frac{1}{6}, 6\right \}$

Correct answer:

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

Explanation:

Rewrite by taking advantage of the Product of Powers Property and the Power of a Power Property:

$6^{2x+1} + 6 = 6 ^{x+2} + 6 ^{x}$

$6^{1 } \cdot 6^{2x} + 6 = 6 ^{2} \cdot 6 ^{x } + 6 ^{x}$

$6 \cdot 6^{2x} + 6 =36 \cdot 6 ^{x } + 6 ^{x}$

$6 \cdot (6^{x } )^{2 }+ 6 = 37 \cdot( 6 ^{x })$

Substitute for $6^{x}$ ; the resulting equation is the quadratic equation

$6 t ^{2 }+ 6 = 37t$

which can be written in standard form by subtracting 37t from both sides:

$6 t ^{2 }+ 6 - 37t = 37t - 37t$

$6 t ^{2 }- 37t + 6 = 0$

The trinomial can be factored by the method, Look for two integers with sum and product $6 \cdot 6 = 36$ ; by trial and error, we find they are -1, -36 , so the equation can be rewritten and solved by grouping:

$6 t ^{2 }- 36t - t + 6 = 0$

$(6 t ^{2 }- 36t) - (t - 6) = 0$

6t (t - 6) - 1 (t - 6) = 0

(6t - 1 )(t - 6) = 0

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

Either

6t - 1 = 0

6t - 1 + 1 = 0 + 1

6t = 1

$\frac{6t}{6} = \frac{1}{6}$

$t = \frac{1}{6}$

Substituting $6^{x}$ back for :

$6^{x} = \frac{1}{6}$

$6^{x} = 6^{-1 }$

x = -1

Or:

t- 6 = 0

t- 6 + 6 = 0 + 6

t = 6

Substituting $6^{x}$ back for :

$6^{x} = 6$

$6^{x} = 6^{1}$

x = 1

The solution set, as can be confirmed by substituting in the equation, is $\left \{ -1, 1 \right \}$ .

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Example Question #11 : Exponents And Logarithms

Solve for :

$9 ^{x+6} = 27 ^{2x}$

Possible Answers:

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

x = 3

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

x = 27

x = 4

Correct answer:

x = 3

Explanation:

$9 = 3^{2}$ and $27 = 3 ^{3}$ , so,

$9 ^{x+6} = 27 ^{2x}$

can be rewritten as

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

Applying the Power of a Power Rule,

$3 ^{2 \cdot (x+6)} = 3 ^{3 \cdot 2x}$

$3 ^{2 x+12} = 3 ^{6x}$

2x+12 = 6x

2x+12 - 2x = 6x - 2x

12 = 4x

$\frac{12 }{4}= \frac{4x}{4}$

x = 3

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Example Question #11 : Exponents And Logarithms

Solve the equation: $3^9 = 9^{3x-3}$

Possible Answers:

$\frac{5}{2}$

$\frac{5}{3}$

Correct answer:

$\frac{5}{2}$

Explanation:

Rewrite the base of the right side.

9=3^2

$3^9 = 3^{2(3x-3)}$

9=2(3x-3)

Simplify the right side.

9= 6x-6

Add 6 on both sides.

9+6= 6x-6+6

15 = 6x

Divide by 6 on both sides.

$\frac{15}{6} =\frac{ 6x}{6}$

The answer is: $\frac{5}{2}$

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Example Question #12 : Exponents And Logarithms

Solve: $8^3 = 2^{2x}$

Possible Answers:

$\frac{5}{2}$

$\frac{9}{4}$

$\frac{9}{2}$

Correct answer:

$\frac{9}{2}$

Explanation:

Change the base of the left side to base two.

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

The equation becomes:

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

Set the exponents equal since they have similar bases.

9=2x

Divide by 2 on both sides.

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

The answer is: $\frac{9}{2}$

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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 #1 : Exponents And Logarithms

Example Question #2 : Exponents And Logarithms

Example Question #5 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

Example Question #5 : Exponents And Logarithms

Example Question #41 : Sat Subject Test In Math Ii

Example Question #41 : Sat Subject Test In Math Ii

Example Question #11 : Exponents And Logarithms

Example Question #11 : Exponents And Logarithms

Example Question #12 : Exponents And Logarithms

All SAT II Math II Resources

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