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

⇒ 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
4

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.
5

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

Answer:

-5

6

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
7

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
8

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

Answer:

8

9

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.
10

Question text not available

Answer:

Insufficient information

11

Both are extraneous solutions.Only 3 is an extraneous solution. Only –3 is an extraneous solution. Neither is an extraneous solution.

A
Both are extraneous solutions.
B
Only 3 is an extraneous solution.
C
Only –3 is an extraneous solution.
D
Neither is an extraneous solution.
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

Which equation can you solve to find the potential solutions to the equation log2x + log2(x – 6) = 4?

A
x2 – 6x – 4 = 0
B
x2 – 6x – 8 = 0
C
x2 – 6x – 16 = 0
15

What is the solution to log7(x – 4) = log7(4x + 5) ?

A
x = –3
B
x = –2
C
x = –1
D
x = 0
E
There is no solution.

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