USE THE ONE TO ONE PROPERTY OF LOGARITHMS TO SOLVE

Use the One-to-One Property to solve the equation for x.

Problem 1 :

log₂(x + 1) = log₂ 4

Solution:

log₂(x + 1) = log₂ 4

Use One to One Property,

x + 1 = 4

x = 4 - 1

x = 3

Problem 2 :

log₂(x - 3) = log₂ 9

Solution:

log₂(x - 3) = log₂ 9

Use One to One Property,

x - 3 = 9

x = 9 + 3

x = 12

Problem 3 :

log(2x + 1) = log 15

Solution:

log(2x + 1) = log 15

Use One to One Property,

2x + 1 = 15

2x = 14

x = 14/2

x = 7

Problem 4 :

log(5x + 3) = log 12

Solution:

log(5x + 3) = log 12

Use One to One Property,

5x + 3 = 12

5x = 9

x = 9/5

Problem 5 :

ln(x + 2) = ln 6

Solution:

ln(x + 2) = ln 6

Use One to One Property,

x + 2 = 6

x = 6 - 2

x = 4

Problem 6 :

ln(x - 4) = ln 2

Solution:

ln(x - 4) = ln 2

Use One to One Property,

x - 4 = 2

x = 2 + 4

x = 6

Problem 7 :

ln(x² - 2) = ln 23

Solution :

ln(x² - 2) = ln 23

Use One to One Property,

x² - 2 = 23

x² = 25

x = ±5

Problem 8 :

ln(x² - x) = ln 6

Solution:

ln(x² - x) = ln 6

Use One to One Property,

x² - x = 6

x² - x - 6 = 0

(x - 3) (x + 2) = 0

x = 3 or x = -2

Problem 9 :

A population of 30 mice is expected to double each year. The number p of mice in the population each year is given by p = 30(2ⁿ). In how many years will there be 960 mice in the population?

Solution:

p = 30(2ⁿ)

Number of mice = 960

960 = 30(2ⁿ)

960/30 = 2ⁿ

32 = 2ⁿ

2⁵ = 2ⁿ

n = 5

So, the required number of years is 5.

Problem 10 :

Approximate the solution of each equation using the graph

1 − 5^{5 − x} = −9

Solution :

1 − 5^{5 − x} = −9

1 + 9 = 5^{5 − x}

10 = 5^{5 − x}

log₅10 = 5 - x

log 10 / log 5 = 5 - x

1/0.6989 = 5 - x

1.43 = 5 - x

x = 5 - 1.43

x = 3.56

By observing the graph, the point of intersection of the horizontal line and the curve are the same.

Problem 11 :

Approximate the solution of each equation using the graph

log₂5x = 2

Solution :

log₂5x = 2

5x = 2²

5x = 4

x = 4/5

x = 0.8

By observing the graph, they are the same.

Problem 12 :

The apparent magnitude of a star is a measure of the brightness of the star as it appears to observers on Earth. The apparent magnitude M of the dimmest star that can be seen with a telescope is

M = 5 log D + 2

where D is the diameter (in millimeters) of the telescope’s objective lens. What is the diameter of the objective lens of a telescope that can reveal stars with a magnitude of 12?

Solution :

M = 5 log D + 2

Magnitude = 12

12 = 5 log D + 2

12 - 2 = 5 log D

10 = 5 log D

10/5 = log D

2 = log D

10² = D

D = 100

So, the required diameter is 100 millimeter.

Problem 13 :

A biologist can estimate the age of an African elephant by measuring the length of its footprint and using the equation

ℓ = 45 − 25.7e^−0.09a

where ℓ is the length (in centimeters) of the footprint and a is the age (in years).

a. Rewrite the equation, solving for a in terms of ℓ.

b. Use the equation in part (a) to find the ages of the elephants whose footprints are shown.

Solution :

ℓ = 45 − 25.7e^−0.09a

a) Solving for a :

25.7e^−0.09a = 45 - l

e^−0.09a = (45 - l)/25.7

-0.09a = ln [(45 - l) / 25.7]

-0.09a = ln (45 - l) - ln 25.7

0.09a = ln 25.7 - ln (45 - l)

0.09 a = ln [25.7/(45 - l)]

a = (1/0.09) ln [25.7/(45 - l)]

b)

Age of elephant whose length of foot print = 36 cm

a = (1/0.09) ln [25.7/(45 - 36)]

= 11.11 ln (25.7/9)

= 11.1 (1.04)

= 11.54

approximately 12 years old.

Age of elephant whose length of foot print = 32 cm

a = (1/0.09) ln [25.7/(45 - 32)]

= 11.11 ln (25.7/13)

= 11.1 ln (1.97)

= 11.1 (0.68)

= 7.54

approximately 8 years old.

Age of elephant whose length of foot print = 28 cm

a = (1/0.09) ln [25.7/(45 - 28)]

= 11.11 ln (25.7/17)

= 11.1 ln (1.51)

= 11.1 (0.413)

= 4.58

approximately 5 years old.

Age of elephant whose length of foot print = 24 cm

a = (1/0.09) ln [25.7/(45 - 24)]

= 11.11 ln (25.7/21)

= 11.1 ln (1.22)

= 11.1 (0.20)

= 2.24

approximately 2 years old.