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Find the value of y.
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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 |
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 β 219eβ0.05t
where t is the age (in years) of the shark. How old is a shark that is 175 centimeters long?
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 :
log7 3 + log7 x = log7 32
Problem 2 :
2 log6 4x = 0
Problem 3 :
log2 x + log2 (x - 3) = 2
Problem 4 :
log2 (x + 5) - log2 (x - 2) = 3
Problem 5 :
4 ln (2x + 3) = 11
Problem 6 :
log x - log 6 = 2 log 4
Problem 7 :
log 2x = 1.5
Problem 8 :
log2 2x = -0.65
Problem 9 :
1/3 log2 x + 5 = 7
Problem 10 :
4 log5 (x + 1) = 4.8
Problem 11 :
log2 x + log2 3 = 3
Problem 12 :
2 log4 x - log4 (x - 1) = 1
Solve the equation. Check for extraneous solutions.
Problem 13 :
ln (7x β 4) = ln (2x + 11)
Problem 14 :
log2(x β 6) = 5
Problem 15 :
log 5x + log (x β 1) = 2
Problem 16 :
log4(x + 12) + log4 x = 3
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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
Problem 2 :
log(3x - 2) = 2
Problem 3 :
2 logx = log2 + log(3x - 4)
Problem 4 :
log x + log(x - 1) = log(4x)
Problem 5 :
log3(x + 25) - log3(x - 1) = 3
Problem 6 :
log9(x - 5) + log9(x + 3) = 1
Problem 7 :
log x + log(x - 3) = 1
Problem 8 :
log2(x - 2) + log2(x + 1) = 2
Problem 9 :
Given that
2 log3(x - 5) - log3(2x - 13) = 1
Show that x2 - 16x + 64 = 0 and solve for x.
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
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.
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 = 3β3 and a = -3β3, b = β3 and -β3
Use the One-to-One Property to solve the equation for x.
Problem 1 :
log2(x + 1) = log2 4
Problem 2 :
log2(x - 3) = log2 9
Problem 3 :
log(2x + 1) = log 15
Problem 4 :
log(5x + 3) = log 12
Problem 5 :
ln(x + 2) = ln 6
Problem 6 :
ln(x - 4) = ln 2
Problem 7 :
ln(x2 - 2) = ln 23
Problem 8 :
ln(x2 - x) = ln 6
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?
Problem 10 :
Approximate the solution of each equation using the graph
1 β 55 β x = β9

Problem 11 :
Approximate the solution of each equation using the graph
log25x = 2

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 telescopeβs objective lens. What is the diameter of the objective lens of a telescope that can reveal stars with a magnitude of 12?
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 β 25.7eβ0.09a
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.

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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May 21, 24 08:51 PM
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