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A2.5.D Solve … single logarithmic equations having real solutions. Also A2.5.E
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Explore Solving Logarithmic Equations GraphicallyOne way to solve logarithmic equations is graphically. First, graph each side of the equation separately. The point(s) at which the two graphs intersect are the solutions of the equation.
Look at the equation 12.2 + 5.45 ln x = 12.5 + 5.2 ln x. To solve the equation graphically, split it into two separate equations.
y 1 =
y 2 =
What will the graphs of y 1 and y 2 look like?
Graph y 1 and y 2 using a graphing calculator.
The x-coordinate of the point of intersection is approximately .
So, the solution of the equation is x ≈ .
Reflect
1. How can you check the solution of a logarithmic equation after it is found graphically?
2. How would you graph and solve a logarithmic equation where the base of the logarithmic function is not 10 or e if your calculator only graphed those bases?
Module 16 899 Lesson 3
16 . 3 Solving Logarithmic EquationsEssential Question: What are some ways you can solve logarithmic equations?
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Explain 1 Solving Logarithmic Equations AlgebraicallyIn addition to solving logarithmic equations graphically, logarithmic equations can be solved algebraically. The inverse relationship between logarithmic and exponential functions allows you to rewrite log b x = a as b a = x.
Example 1 Solve the equations. Check for extraneous solutions.
7 + log 3 (5x - 4) = 10
7 + log 3 (5x - 4) = 10 Original equation
7 - 7 + log 3 (5x - 4) = 10 - 7 Subtract 7 from both sides.
log 3 (5x - 4) = 3 Simplify.
3 3 = 5x - 4 Definition of a logarithm
27 = 5x - 4 Evaluate the exponent.
31 = 5x Add 4 to both sides.
6.2 = x Divide both sides by 5.
Check:
7 + log 3 (5 (6.2) - 4) = 10
7 + log 3 (31 - 4) = 10
7 + log 3 (27) = 10
7 + log 27
_ log 3
= 10
7 + 3 = 10
10 = 10
The solution is x = 6.2.
log x + log (x + 9) = 1
log x + log (x + 9) = 1 Original equation
log (x (x + 9) ) = 1 Product Property of Logarithms
log ( ) = 1 Multiply.
1
= x 2 + 9x Definition of a logarithm
= x 2 + 9x Evaluate the exponent.
= x 2 + 9x - Subtract 10 from both sides.
0 = (x + ) ( x - ) Factor.
x = or x = Solve.
Module 16 900 Lesson 3
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Check:
log (-10) + log (-10 + 9) = 1 log 1 + log (1 + 9) = 1
log ( ) + log ( ) = 1 log 1 + log = 1
= 1 + = 1
x = is an extraneous solution. = 1
The solution is x = .
Reflect
3. Explain how you would solve the equation log 5 45x = 1 + log 5 3. Then solve.
Your Turn
Solve the equations. Check for extraneous solutions.
4. log 4 (2x + 12) + 5 = 8
Module 16 901 Lesson 3
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Your Turn
5. log 6 (x - 5) = 2 - log 6 x
Explain 2 Solving a Real-World Problem Using a Logarithmic Equation
Given a logarithmic function ƒ (x) that models a real-world situation, find the value of x for which ƒ (x) = c, where c is a constant. Check the reasonableness of the solution by graphing.
Example 2 Solve using properties of logarithms. Then check the reasonableness of the solution by graphing.
The energy released by an earthquake can be a very large number, so a much smaller number is reported as the magnitude of the earthquake. The magnitude M of an earthquake with energy E (in ergs) is calculated using the formula M = 2 _ 3 log E - 2.9. The largest earthquake known to have occurred in Texas happened on August 16, 1931 and had a magnitude of 5.8. How much energy did the earthquake have?
Let M = 5.8. Substitute the value for M into the formula for the magnitude and solve for E.
5.8 = 2 _ 3 log E - 2.9 Substitute.
5.8 + 2.9 = 2 _ 3 log E - 2.9 + 2.9 Add 2.9 to both sides.
Module 16 902 Lesson 3
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8.7 = 2 _ 3 log E Simplify.
13.05 = log E Multiply both sides by 3 _ 2 .
E ≈ 10 13.05 Definition of a logarithm
E ≈ 11,200,000,000,000 Evaluate the exponent
The earthquake had about 11,200,000,000,000 ergs of energy.
Check the reasonableness of the solution by graphing both sides of the equation on a graphing calculator.
The two graphs intersect at approximately x = 11,200,000,000,000, so the answer is reasonable.
An earthquake that occurred on December 7, 2013, near Volcano, Hawaii, had a magnitude of 2.9. How much energy did the earthquake have?
Let M = 2.9. Substitute the value for M into the formula for the magnitude and solve for E.
= 2 _ 3 log E - 2.9 Substitute.
2.9 + = 2 _ 3 log E - 2.9 + Add 2.9 to both sides.
= 2 _ 3 log E Simplify.
= log E Multiply both sides by 3 _ 2 .
E = Definition of a logarithm
E ≈ Evaluate the exponent.
The earthquake had about ergs of energy.
Check the reasonableness of the solution by graphing both sides of the equation on a graphing calculator.
The two graphs intersect at approximately x = , so the answer is reasonable.
Module 16 903 Lesson 3
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Reflect
6. Compare the energies of the two earthquakes.
Your Turn
7. The difference between the apparent magnitude (brightness) m of a star and its absolute magnitude M is given by the formula m - M = 5log d __ 10 , where d is the distance of the star from Earth, measured in parsecs. The star Antares has an apparent magnitude of 1.0 and an absolute magnitude of -5.3. Find the distance of Antares from Earth. Check the reasonableness of your answer by graphing.
Elaborate
8. Explain why it’s important to check the solutions of a logarithmic equation.
9. Describe how to solve an exponential equation graphically.
10. Essential Question Check-In Describe how to solve a logarithmic equation algebraically.
Module 16 904 Lesson 3
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• Online Homework• Hints and Help• Extra Practice
Evaluate: Homework and Practice
Solve the equations graphically.
1. 8 + log x = 2 log x - 12
2. log 6 2x = 2log 6 x + 1
3. 10 = 3log 2 x
4. 9.4 - log 5 x = 4log 5 x + 0.5
Module 16 905 Lesson 3
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5. 0.2ln x + 5 = ln x - 6
6. 8log 3 x - 1 _ 4 = 3 _ 4 log 3 x + 2
Solve the equations. Check for extraneous solutions.
7. log 9 (4x + 5) = 2 8. ln 8x = ln 2 + 5
Module 16 906 Lesson 3
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9. log 2 x + log 2 (x - 4) = 5
10. log 6 x = - ( log 6 (x - 1 _ 4 ) + 2)
Module 16 907 Lesson 3
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11. 2.4log 4 x = log 4 3x + 1
12. log 2x + log (x - 5) = 2
13. lo g 5 (4x + 6) = l og 5 (8x - 2) 14. 2ln x - 0.4 = 2.5 - 5ln x
Module 16 908 Lesson 3
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Solve using properties of logarithms. Then check the reasonableness of the solution by graphing.
15. Photography On many cameras, the amount of light admitted through the lens can be controlled by changing the size of the opening, or aperture. The size of the aperture is measured as an f-stop setting. The relationship between the f-stop and the amount of light admitted can be represented by the equation n = log 2 1 __ ℓ , where n is the change in f-stop setting from the starting value, f ___ 5.6 . Solve the equation for ℓ when the f-stop setting is increased to f __ 16 .
F–stop Setting f _ 2
f _ 2.8
f _ 4
f _ 5.6
f _ 8
f _ 11
f _ 16
Change in F–stop Setting -3 -2 -1 0 1 2 3
16. Astronomy A telescope’s limiting magnitude m is the brightness of the faintest star that can be seen using the telescope. The limiting magnitude depends on the diameter d (in millimeters) of the telescope’s objective lens. What diameter lens would be needed to view a faintest star with a brightness of 21?
Formulas for Determining Limiting Magnitude from Lens Diameter
Standard formula m = 2.7 + 5log d
Module 16 909 Lesson 3
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17. The equation for finding pH levels is pH = -log ⎡ ⎣ H + ⎤ ⎦ , where H + is the hydrogen ion concentration. Cow’s milk has a pH level of 6.7 and goat’s milk has a pH of 6.48. Find the difference in the hydrogen ion concentration between the two types of milk.
18. The sound level at a rock concert is 115 decibels. The loudness L of sound in decibels is given by the equation L = 10log ( ℓ __ ℓ 0
) , where I is the intensity of sound and ℓ 0 is the audible sound. If ℓ 0 = 10 -12 decibels, what is the intensity of the sound at the rock
concert?
Module 16 910 Lesson 3
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19. The magnitude M of an earthquake with energy E (in ergs) is calculated using the formula M = 2 __ 3 log E - 2.9. How much energy does an earthquake with a magnitude of 3.8 have?
20. The brightness m of a star is given by the formula m = 5log d __ 10 - 1.6, where d is the distance of the star from Earth, measured in parsecs. Find the distance the star is from Earth if the brightness of the star is 3.2.
21. Match the equations with the solutions.
A. 2lo g 5 x = 10 ――――― x = 390,625
B. lo g 5 2x = 10 ――――― x = 3125
C. lo g 5 x + 2 = 10 ――――― x = 1562.5
D. 2lo g 5 2x = 10 ――――― x = 4,882,812.5
Module 16 911 Lesson 3
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H.O.T. Focus on Higher Order Thinking
22. Multi-Step Charles collected data on the atmospheric pressure (ranging from 4 to 15 pounds per square inch [psi]) and the corresponding altitude above the surface of Earth (ranging from 1 to 30,000 feet). He used exponential and linear regression to write two functions that give the altitude above the surface of the Earth given the atmospheric pressure, x.
ƒ (x) = 66,990 - 24,747ln x
g (x) = -2870x + 40,393
a. At what atmospheric pressure (s) do the equations give the same altitude?
b. At what altitude (s) above Earth do these atmospheric pressure(s) occur?
Module 16 912 Lesson 3
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23. Draw Conclusions Solve the equation lo g 2 (7x + 1) = log 2 (2 - x) by subtracting the logarithms. Is there an easier way to solve this equation? Explain.
24. Explain the Error A student found the solutions of the equation log 6 x + log 6 (x + 5) = 2 to be x = 4 and x = -9. Explain the error in the student’s reasoning.
Module 16 913 Lesson 3
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A telescope’s limiting magnitude m is the brightness of the faintest star that can be seen using the telescope. The limiting magnitude depends on the diameter d (in millimeters) of the telescope’s objective lens. Limiting magnitude can be calculated in different ways. The table gives two formulas relating m to d. One is a standard formula used in astronomy. The other is a proposed new formula based on data gathered from users of telescopes of various lens diameters. For what lens diameter do the two formulas give the same limiting magnitude? Solve by graphing.
Formulas for Determining Limiting Magnitude from Lens Diameter
Standard formula m = 2.7 + 5logd
Proposed formula m = 4.5 + 4.4logd
Lesson Performance Task
Module 16 914 Lesson 3
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