AnswersDHS Algebra 2 VSolving Equations using Properties of Logarithms

Solving Equations using Properties of Logarithms Answers

20 verified answers
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

Question text not available

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

Question text not available

Answer:

5

5

Question text not available

Answer:

1

6

Question text not available

Answer:

4

7

Question text not available

Answer:

3

8

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

Answer:

-5

9

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.
10

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
11

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
12

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
13

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

Answer:

8

14

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.
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.
16

Question text not available

Answer:

3

17

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.
18

⇒ -3

Answer:

-3

19

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 in the solution is treating the logarithms as if they have the same base. The first term, log x, is a common logarithm (base 10), while the other terms have a base of 5. The one-to-one property of logarithms only applies when the bases are identical. To solve the equation correctly, you must first use the change-of-base formula to rewrite the logarithms so they have the same base before attempting to equate the arguments.

20

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.

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