SOLVING LOGARITHMIC FUNCTION WITH BASE WORKSHEET

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Find the value of y.

1)  log5 25 = y

2)  log31 = y

3)  log16 4 = y

4)  log2 (1/8) = y

5)  log51 = y

6)  log2 8 = y

7)  log7 (1/7) = y

8)  log3 (1/9) = y

9)  logy 32 = 5

10)  log9 y = -1/2

11)  log4 (1/8) = y

12)  log9 (1/81) = y

Solution

Problem 13 :

Describe the similarities and difference between in solving the equations 

45x - 2 = 16 and log4(10x + 6) = 1

Then solve the each equation

Problem 14 :

For a sound with intensity I (in watts per square meter) the loudness L(I) of the sound (in decibels) is given by the function

L(I) = 10 log (I/I0)

Where I0 is the intensity of barely audible sound (about 10-12 watts per square meter) An artist in a recording studio turns up the volume of a track so that the intensity of the sound doubles. By how many decibels does the loudness increase ?

Problem 15 :

The length β„“ (in centimeters) of a scalloped hammerhead shark can be modeled by the function

β„“ = 266 βˆ’ 219eβˆ’0.05t

where t is the age (in years) of the shark. How old is a shark that is 175 centimeters long?

Answer Key

1)  y = 2

2)  y = 0

3)  y = 1/2

4)  y = -3

5)  y = 0

6)  y = 3

7)  y = -1

8)  y = -2

9)  y = 2

10)  y = 1/3

11)  y = -3/2

12)  y = -2

13) x = -1/5

14) The loudness increases by 10 log 2 decibels or about 3 decibels.

15) Approximately 18 years.

Problem 1 :

log3 + logx = log32

Solution

Problem 2 :

2 log6 4x = 0

Solution

Problem 3 :

logx + log2 (x - 3) = 2

Solution

Problem 4 :

log(x + 5) - log2 (x - 2) = 3

Solution

Problem 5 :

4 ln (2x + 3) = 11

Solution

Problem 6 :

log x - log 6 = 2 log 4

Solution

Problem 7 :

log 2x = 1.5

Solution

Problem 8 :

log2x = -0.65

Solution

Problem 9 :

1/3 log2 x + 5 = 7

Solution

Problem 10 :

4 log(x + 1) = 4.8

Solution

Problem 11 :

log2 x + log2 3 = 3

Solution

Problem 12 :

2 log4 x - log(x - 1) = 1

Solution

Solve the equation. Check for extraneous solutions.

Problem 13 :

ln (7x βˆ’ 4) = ln (2x + 11)

Solution

Problem 14 :

log2(x βˆ’ 6) = 5

Solution

Problem 15 :

log 5x + log (x βˆ’ 1) = 2

Solution

Problem 16 :

log4(x + 12) + log4 x = 3

Solution

Answer Key

1)  x = 32/3

2)  x = 1/4

3)  x = 4

4)  x = 3

5)  x = 6.321

6)  x = 96

7)  x = 15.81

8)  x = 0.32

9)  x = 26

10)  x = 5.90

11)  x = 8/3

12)  x = 2

13) x = 3

14) x = 38

15) x = -4 and x = 5

16) x = 4 and x = -16

Solve the following logarithmic equations.

Problem 1 :

ln x = -3

Solution

Problem 2 :

log(3x - 2) = 2

Solution

Problem 3 :

2 logx = log2 + log(3x - 4)

Solution

Problem 4 :

log x + log(x - 1) = log(4x)

Solution

Problem 5 :

log3(x + 25) - log3(x - 1) = 3

Solution

Problem 6 :

log9(x - 5) + log9(x + 3) = 1

Solution

Problem 7 :

log x + log(x - 3) = 1

Solution

Problem 8 :

log2(x - 2) + log2(x + 1) = 2

Solution

Problem 9 :

Given that 

log3(x - 5) - log3(2x - 13) = 1

Show that x2 - 16x + 64 = 0 and solve for x.

Solution

Problem 10 :

a) Find the positive value of x such that 

log x64 = 2

b) Solve for x 

log2(11 - 6x) = 2log2(x - 1) + 3

Solution

Problem 11 :

Given that a and b are positive constants, solve the simultaneous equations

a = 3b

log3 a + log3 b = 2

Give your answers as exact numbers.

Solution

Answer Key

1) x = e-3

2) x = 34

3) x = 4, x = 2

4) x = 0, x = 5

5) x = 2

6) x = 6, x = -4

7) x = 5 or x = -2

8) x = 3 or x = -2

9) x = 8 and x = 8

10) x = 8, x = -1/4 and x = 3/2

11) a = 3√3 and a = -3√3, b = βˆš3 and -√3

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

Problem 1 :

log2(x + 1) = log4

Solution

Problem 2 :

log2(x - 3) = log9

Solution

Problem 3 :

log(2x + 1) = log 15

Solution

Problem 4 :

log(5x + 3) = log 12

Solution

Problem 5 :

ln(x + 2) = ln 6

Solution

Problem 6 :

ln(x - 4) = ln 2

Solution

Problem 7 :

ln(x2 - 2) = ln 23

Solution

Problem 8 :

ln(x2 - x) = ln 6

Solution

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

Problem 10 :

Approximate the solution of each equation using the graph

1 βˆ’ 55 βˆ’ x = βˆ’9

solving-log-equation-q1.png

Solution

Problem 11 :

Approximate the solution of each equation using the graph

log25x = 2

solving-log-equation-q2.png

Solution

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

Problem 13 :

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

β„“ = 45 βˆ’ 25.7eβˆ’0.09a

where β„“ 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 β„“.

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

solving-log-equation-q3.png

Solution

Answer Key

1) x = 3

2) x = 12

3) x = 7

4) x = 9/5

5) x = 4

6) x = 6

7) x = Β±5

8) x = 3 or x = -2

9) the required number of years is 5.

10) x = 3.56

11) x = 0.8

12) the required diameter is 100 millimeter.

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

b) Age of elephant whose length of foot print = 36 cm, 12 years old.

Age of elephant whose length of foot print = 32 cm, 8 years old

Age of elephant whose length of foot print = 28 cm, 5 years old

Age of elephant whose length of foot print = 24 cm, 2 years old.

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