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SAT II Math II : Exponents and Logarithms

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

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Example Questions

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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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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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SAT II Math II : Exponents and Logarithms

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

All SAT II Math II Resources

Example Questions

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

Example Question #21 : Exponents And Logarithms

Example Question #21 : Exponents And Logarithms

Example Question #21 : Exponents And Logarithms

All SAT II Math II Resources

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