AnswersDHS Algebra 2 VSolving Logarithmic Equations using Technology

Properties of Logarithms Answers

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1
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Kim solved the equation below by graphing a system of equations.What is the approximate solution to the equation?

Question illustration
A
0.6
B
0.9
C
1.4
D
1.6
2
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Omar wants to use a graph to solve the equation below.Which system of equations should Omar use?

Question illustration
A
Option
Option A
B
Option
Option B
C
Option
Option C
D
Option
Option D
3

Which is equivalent to ?

Question illustration
A
Option
Option A
B
Option
Option B
C
Option
Option C
D
Option
Option D
4

Kim solved the equation below by graphing a system of equations.What is the approximate solution to the equation?

Question illustration
A
0.6
B
0.9
C
1.4
D
1.6
5

Which system of equations could be graphed to solve the equation below?

Question illustration
A
y 1 = StartFraction log 0.5 Over x EndFraction , y 2 = StartFraction log 3 Over 2 + x EndFraction
Option A
B
y 1 = StartFraction log x Over log 0.5 EndFraction, y 2 = StartFraction log 2 + x Over log 3 EndFraction
Option B
C
y 1 = StartFraction log 0.5 Over log 3 EndFraction, y 2 = StartFraction log x Over log 2 + x EndFraction
Option C
D
y 1 = StartFraction log x Over log 0.5 EndFraction, y 2 = StartFraction log 2 Over log 3 EndFraction + x
Option D
7

Tyler applied the change of base formula to a logarithmic expression. The resulting expression is shown below. Which expression could be Tyler’s original expression?

Question illustration
A
Log Subscript one-fourth Baseline 12
Option A
B
Log Subscript 12 Baseline One-fourth
Option B
C
12 Log one-fourth
Option C
D
One-fourth log 12
Option D
8

What is the approximate value of q in the equation below?

Question illustration
A
–1.613
B
–1.522
C
0.585
D
3.079
9

Which system of equations could be graphed to solve the equation below?

Question illustration
A
y 1 = 3 x, y 2 = 2 x
Option A
B
y 1 = log (2 x + 1), y 2 = 3 x minus 2
Option B
C
y 1 = log 2 x + 1, y 2 = 3 x minus 2
Option C
D
y 1 = log (2 x + 1 + 2), y 2 = 3 x
Option D
10

Devonte used the change of base formula to approximate . Which expression did Devonte use?

Question illustration
A
Log StartFraction 8 Over 25 EndFraction
Option A
B
StartFraction log 8 Over log 25 EndFraction
Option B
C
log StartFraction 25 Over 8 EndFraction
Option C
D
StartFraction log 25 Over log 8 EndFraction
Option D
11

Which expression can be used to approximate the expression below, for all positive numbers a, b, and x, where a 1 and b 1?

Question illustration
A
Option
Option A
B
Option
Option B
C
Option
Option C
D
Option
Option D
12

Tenisha solved the equation below by graphing a system of equations.Which point approximates the solution for Tenisha’s system of equations?

Question illustration
A
(0.9, 0.8)
B
(1.0, 1.4)
C
(2.3, 1.1)
D
(2.7, 13.3)
13

Which is equivalent to ?

Question illustration
A
log n = StartFraction log 2 Over 4 EndFraction
Option A
B
n = StartFraction log 2 Over log 4 EndFraction
Option B
C
n = log 4 times log 2
Option C
D
log n = 4 log 2
Option D
14

What is the approximate value of q in the equation below?

Question illustration
–
–1.613
–
–1.522
0
0.585
3
3.079
15

What is the approximate value of x in the equation below?

Question illustration
A
–2.523
B
–1.317
C
2.880
D
7.485
16

Consider the equation below.Which system of equations can represent the equation?

Question illustration
A
y 1 = StartFraction log (x + 3) Over log 4 EndFraction, y 2 = StartFraction log (2 + x) Over log 2 EndFraction
Option A
B
y 1 = StartFraction log x + 3 Over log 4 EndFraction, y 2 = StartFraction log 2 + x Over log 2 EndFraction
Option B
C
y 1 = StartFraction log 4 Over log 2 EndFraction, y 2 = StartFraction log (x + 3) Over log (2 + x) EndFraction
Option C
D
y 1 = StartFraction log x + 3 Over 4 EndFraction, y 2 = StartFraction log 2 + x Over 2 EndFraction
Option D
17

A teacher used the change of base formula to determine whether the equation below is correct.Which statement explains whether the equation is correct?

Question illustration
A
The equation is correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = log (2 times 10) times log (4 times 8) times log (10 times 4). = log (20) times log (32) times log (40). =3
Option A
B
The equation is correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = StartFraction log 10 Over log 2 EndFraction times StartFraction log 8 Over log 4 EndFraction times StartFraction log 4 Over log 10 EndFraction. = StartFraction log 8 Over log 2 EndFraction. = 3
Option B
C
The equation is correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = StartFraction log 10 Over log 2 EndFraction times StartFraction log 8 Over log 4 EndFraction times StartFraction log 4 Over log 10 EndFraction. = StartFraction log 8 Over log 2 EndFraction. = 4.
Option C
D
The equation is not correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = log StartFraction 10 Over 2 EndFraction times log eight-fourths times log four-tenths. = log 5 times log 2 times log 0.4. = negative 0.08
Option D
18

What is the approximate value of x in the equation below.

Question illustration
A
–3.396
B
–0.708
C
0.304
D
0.955
19

Tenisha solved the equation below by graphing a system of equations.Which point approximates the solution for Tenisha’s system of equations?

Question illustration
A
(0.9, 0.8)
B
(1.0, 1.4)
C
(2.3, 1.1)
D
(2.7, 13.3)
20

Which expression results when the change of base formula is applied to ?

Question illustration
A
StartFraction log (x + 2) Over log 4 EndFraction
Option A
B
StartFraction log 4 Over log (x + 2) EndFraction
Option B
C
StartFraction log 4 Over log x + 2 EndFraction
Option C
D
StartFraction log x + 2 Over log 4 EndFraction
Option D
21

Omar wants to use a graph to solve the equation below.Which system of equations should Omar use?

Question illustration
A
y 1 = StartFraction log 6 Over x EndFraction , y 2 = StartFraction log 2 Over x + 4 EndFraction
Option A
B
y 1 = StartFraction log x Over log 6 EndFraction, y 2 = StartFraction log x + 4 Over log 2 EndFraction
Option B
C
y 1 = StartFraction log 6 Over log 2 EndFraction, y 2 = StartFraction log x Over log (x + 4) EndFraction
Option C
D
y 1 = StartFraction log x Over log 6 EndFraction, y 2 = StartFraction log (x + 4) Over log 2 EndFraction
Option D

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