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High School Math : Algebra II

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Example Questions

Example Question #171 : Algebra Ii

Solve for :

$2 + \log_{3} (x - 2) = \log_{3} (x +6)$

Possible Answers:

The equation has no solution.

x = 6

$x = 6 \textrm{ or }x=10$

x = 10

x = 3

Correct answer:

x = 3

Explanation:

Since $3^{2} = 9$ , $2 = \log_{3}9$ , and we can rewrite and solve this statement as follows:

$2 + \log_{3} (x - 2) = \log_{3} (x +6)$

$\log_{3}9 + \log_{3} (x - 2) = \log_{3} (x +6)$

$\log_{3} 9 (x - 2) = \log_{3} (x +6)$

$3 ^ { \log_{3} 9 (x - 2)} =3 ^ { \log_{3} (x +6)}$

9(x-2) = x + 6

9x-18 = x + 6

9x-18 -x + 18 = x + 6 -x + 18

8x = 24

x = 3

Substitution confirms this to be the only solution.

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Example Question #22 : Logarithms

Solve for :

$\log_{4} (x-3) = \log_{16} (x-1)$

Possible Answers:

$x = 2 \textrm{ or } x = 5$

The equation has no solution.

The solution set of the equation is the set of all real numbers.

x = 5

x = 2

Correct answer:

x = 5

Explanation:

We can rewrite this as follows:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (x-3) =\frac{ \log_{4} (x-1)}{ \log_{4} 16}$

$\log_{4} (x-3) =\frac{ \log_{4} (x-1)}{ 2}$

$2 \log_{4} (x-3) =\log_{4} (x-1)$

$\log_{4} \left [(x-3)^2\right ] =\log_{4} (x-1)$

$\dpi{120} 4^ {\log_{4} \left [(x-3)^2\right ]} =4^ { \log_{4} (x-1)}$

(x-3)^2 = x -1

x^2 - 6x + 9 = x -1

x^2 - 6x + 9-x +1 = x-1 -x +1

x^2 - 7x + 10 = 0

(x-2)(x-5) = 0

Split this into a compound statement, and solve:

$x - 2 \Rightarrow x = 2$

$x - 5 \Rightarrow x = 5$

Test both of these possible solutions by substituting:

x = 2:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (2-3) = \log_{16} (2-1)$

$\log_{4} (-1) = \log_{16} 1$

Since only positive numbers can have logarithms, this equation does not make sense. x = 2 is not a solution.

x = 5:

$\log_{4} (x-3) = \log_{16} (x-1)$

$\log_{4} (5-3) = \log_{16} (5-1)$

$\log_{4} 2 = \log_{16} 4$

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

x = 5 is a solution - the only solution.

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Example Question #173 : Algebra Ii

What is the sum of the value(s) of x that satisfy the equation $\log_{10}(2x^2-35x)=2$ ?

Possible Answers:

No solution.

100

35/2

Correct answer:

35/2

Explanation:

$\log_{10}(2x^2-35x)=2$

We need to rewrite the logarithmic equation into a form that is more recognizable.

An equation in the form $\log_{a}b=c$ can be rewritten in the exponential form b=a^c , where a, b, and c are constants (and b > 0). Thus, we will rewrite our original equation as follows:

2x^2-35x=10^2 .

We can now approach this as we would a typical quadratic equation.

2x^2-35x=10^2=100

Subtract 100 from both sides. We need to set all of the terms equal to zero.

2x^2-35x-100=0 .

We can factor this by first multiplying the outer coefficients. The product of 2 and -100 is equal to -200. We must think of two numbers that multiply to give -200 but add to give -35 (the middle coefficient). Two numbers that satisfy both of these requirements are -40 and 5. We will now rewrite the quadratic polynomial.

2x^2-35x-100=(2x^2-40x)+(5x-100)=0

We can use grouping to factor the polynomial. Factor the first two terms and the last two terms separately.

(2x^2-40x)+(5x-100)=0

2x(x-20)+5(x-20)=0

Notice that we can now factor out x - 20 from the 2x term and the 5 term.

(2x+5)(x-20)=0

To solve this, we set each factor equal to zero.

2x+5=0

$x=\frac{-5}{2}$

x-20=0

x=20.

The question ultimately asks for the sum of the values of x that satisfy the equation, so we must add -5/2 and 20, which yields 35/2.

The answer is 35/2.

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Example Question #15 : Solving Logarithmic Equations

Solve the following logarithmic equation:

$log_{9}(x+4)+log_{9}(x-4)=1$

Possible Answers:

x=-5

x=6

x=1

x=4

x=5

Correct answer:

x=5

Explanation:

Since the bases of the logs are the same and the logarithms are added, the arguments can be multiplied together.

$log_{9}(x+4)+log_{9}(x-4)=1$

$log_{9}(x+4)(x-4)=1$

The logarithm can then be converted to exponential form.

$(x+4)(x-4)=9^{1}$

x^2-16=9

x^2-25=0

(x-5)(x+5)=0

x=5

Although there are two solutions to the equation, logarithims cannot be negative. Therefore, the only real solution is .

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Example Question #25 : Logarithms

Solve the following logarithmic equation:

$log_{2}[\left log_{9}(log_{3}x) \right ]=-1$

Possible Answers:

x=3

x=36

x=18

x=27

x=9

Correct answer:

x=27

Explanation:

Start by converting the logarithms into exponential form:

$log_{2}[\left log_{9}(log_{3}x) \right ]=-1$

$log_{9}(log_{3}x) =2^{-1}$

$log_{9}(log_{3}x) =\frac{1}{2}$

Convert again into exponential form:

$log_{3}x =9^{\frac{1}{2}}$

Simplify and solve:

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

$log_{3}x =3$

x=3^3=27

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Example Question #23 : Logarithms

Solve the following logarithmic equation:

$log_{2}(3x-2)=log_{2}(2x+6)$

Possible Answers:

x=6

x=10

x=5

x=7

x=8

Correct answer:

x=8

Explanation:

Since the bases of the logs are the same, the terms inside the logs can be set equal to each other.

3x-2=2x+6

x=8

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Example Question #13 : Solving Logarithmic Equations

Solve the equation.

$2^{9x-3}=\left(\frac{1}{32}\right)^{x-11}$

Possible Answers:

$x=\frac{29}{7}$

$x=\frac{22}{14}$

$x=\frac{1}{9}$

$x=-\frac{26}{7}$

$x=\frac{22}{7}$

Correct answer:

$x=\frac{29}{7}$

Explanation:

Change the right side to $2^{-5}$ so that both sides are the same. Apply log to both sides so that you can set the exponential expressions equal to each other ( 9x-3=-5x+55 ).

$x=\frac{29}{7}$

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Example Question #171 : Mathematical Relationships And Basic Graphs

What is the value of ?

Possible Answers:

Correct answer:

Explanation:

! is the symbol for factorial, which means the product of the whole numbers less than the given number.

Thus, 4! = 4 * 3 * 2 * 1 = 24 .

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Example Question #1 : Factorials

What is the value of ?

Possible Answers:

$\frac{1}{4}$

Correct answer:

Explanation:

By definition, $4! = 4 \cdot 3 \cdot 2 \cdot 1= 24$ .

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Example Question #3 : Factorials

Find the value of:

$\frac{7!}{3!}$

Possible Answers:

2.33

840

2520

210

Correct answer:

840

Explanation:

The factorial sign (!) just tells us to multiply that number by every integer that leads up to it. So, $\frac{7!}{3!}$ can also be written as:

$\frac{(7\times6\times5\times4\times3\times2\times1)}{(3\times2\times1)}$

To make this easier for ourselves, we can cancel out the numbers that appear on both the top and bottom:

$7\times6\times5\times4 = 840$

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High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #171 : Algebra Ii

Example Question #22 : Logarithms

Example Question #173 : Algebra Ii

Example Question #15 : Solving Logarithmic Equations

Example Question #25 : Logarithms

Example Question #23 : Logarithms

Example Question #13 : Solving Logarithmic Equations

Example Question #171 : Mathematical Relationships And Basic Graphs

Example Question #1 : Factorials

Example Question #3 : Factorials

All High School Math Resources

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