AnswersDHS Algebra 2 VSolving Equations using Properties of Logarithms

Solving Equations using Properties of Logarithms Answers

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1
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Solve the equation log4(x + 20) = 3x =

Answer:

44

2
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Solve the equation log6(13 – x) = 1.x =

Answer:

7

3

The solution to the equation log2x + log2(x – 6) = 4 is x = .

Answer:

8

4

Question text not available

Answer:

4

5

Question text not available

Answer:

3

6

⇒ 2

Answers:
x2 − 2x − 15 = 0:2
Potential solutions are −3 and 5:5
x2 − 15 = 2x:1
x − 5 = 0 or x + 3 = 0:4
(x − 5)(x + 3) = 0:3
7

Find the potential solution to the equationlog4(2 – x) = log4(–5x – 18).x =

Answer:

-5

8

Which is the only solution to the equationlog3(x2 + 6x) = log3(2x + 12)?

A
x = –6
B
x = –2
C
x = 0
D
x = 2
E
x = 6
9

Which shows the first step in the solution to the equation log2x + log2(x – 6) = 4?

A
log2[x(x – 6)] = 4
Option A
B
x + x – 6 = 4
Option B
C
x + x – 6 = 24
Option C
D
Option
Option D
10

What are the solutions tolog3x + log3(x2 + 2) = 1 + 2log3x?

A
x = –2
B
x = –1
C
x = 1
D
x = 2
E
There is no true solution.
11

Question text not available

Answer:

Insufficient information

12

Explain the error in the solution below. What additional step needs to be completed?log x – log53 = 2log53 log x = 3log53 log x = log533 x = 27

Answer:

The error is applying the one-to-one property when the logarithms have different bases. The additional step is to use the change-of-base formula to rewrite the logarithms with the same base before solving.

13

Which of the following did you include in your response?

A
The bases of the logarithms are not the same.
B
The one-to-one property does not apply when the bases are not the same.
C
The change of base formula should have been used to write the logarithms with the same base.
14

x2 − 2x − 15 = 0 Potential solutions are −3 and 5 x2 − 15 = 2x x − 5 = 0 or x + 3 = 0 (x − 5)(x + 3) = 0

Answers:
x2 − 2x − 15 = 0:2
Potential solutions are −3 and 5:5
x2 − 15 = 2x:1
x − 5 = 0 or x + 3 = 0:4
(x − 5)(x + 3) = 0:3
15

Find the potential solution to the equationlog4(2 – x) = log4(–5x – 18).x =

Answer:

-5

16

Which statement is true about the potential solution x = –5?

A
This number is not a true solution because it is negative.
B
This number is not a true solution because –5x – 18 is negative.
C
This number is not a true solution because 2 – x is negative.
D
This number is a true solution of the original equation.
17

Which is the only solution to the equationlog3(x2 + 6x) = log3(2x + 12)?

A
x = –6
B
x = –2
C
x = 0
D
x = 2
E
x = 6
18

Which shows the first step in the solution to the equation log2x + log2(x – 6) = 4?

A
log2[x(x – 6)] = 4
Option A
B
x + x – 6 = 4
Option B
C
x + x – 6 = 24
Option C
D
Option
Option D
19

What are the solutions tolog3x + log3(x2 + 2) = 1 + 2log3x?

A
x = –2
B
x = –1
C
x = 1
D
x = 2
E
There is no true solution.
20

List the potential solutions to 2log x − log3 = log3 from least to greatest. x = or x =

Answers:
Blank 1:-3
Blank 2:3

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