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1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to...

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1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4 . 4. Condense 2log 5 y – 2log 5 5 + ½log 5 x. 5. Use the change of base formula to evaluate log 8 20. 1 Algebra II
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Page 1: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

1. Use a property of logarithms to evaluate log392

2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150.

3. Expand ln 732x4.

4. Condense 2log5y – 2log55 + ½log5x.

5. Use the change of base formula to evaluate log820.

1Algebra II

Page 2: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

Solving Exponential and Logarithmic Equations

Algebra II

Page 3: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

To solve equations you must undo what is being done

Exponential and logarithmic functions are inverse operations

You undo “exponentiating” by ”logarizing” and you undo “logarizing” by “exponentiating”

3Algebra II

Page 4: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

1. ax = ay If x = y2. Logbx = logby if x = yExamples:

If 2x = 23 then x = 3 If ln x = 3 then e lnx = e3, so x = e3 If log3x = log3(5), then x = 5 If ex = 7 then ln ex = ln7 , so x = ln7

4Algebra II

Page 5: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

Algebra II 5

Page 6: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

Algebra II 6

83 xx

82 x4x

1432 xx

2x

Page 7: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

Algebra II 7

xx 314 1x12 x

Page 8: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

8Algebra II

Page 9: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

9Algebra II

2x

42 x

521 log01 log 2 x

152 log2 x2 2

152102 x

Page 10: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

10Algebra II

1xe

0x

44 xe21ln ln 2 xe

12ln 2 x2 2

122 xe

Page 11: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

11Algebra II

133 xx123 x

x22 1x

Page 12: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

12Algebra II

121 xx11 x

2xNo Solution

(Extraneous solution)

Page 13: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

13Algebra II

1144 xx1154 x

x515 3x

Page 14: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

14Algebra II

9x

Page 15: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

15Algebra II

23x

22 log5 x

Page 16: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

16Algebra II

4

52 ln x

Page 17: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

17Algebra II

Check both

solutions!

Page 18: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

18Algebra II

Page 19: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

19Algebra II

Page 20: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

20Algebra II

Page 21: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

21Algebra II

Page 22: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

22Algebra II

Page 23: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

1. ln x = x2 – 2 2. Log3x = x2 – 2

23Algebra II

Page 24: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

The population size y of a community of lemmings varies according to the relationship y = y0e0.15t. In this formula, t

is time in months and y0 is the initial population at the time 0. Estimate the population after 8 months if there

were originally 2000 lemmings.

y = y0e0.15t

y = 2000e0.15(8)

y = 2000e1.2

y ≈ 6640.2339

In 8 months, the population will be approximately 6640 lemmings.

Algebra II 24

Page 25: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

How long does it take an investment of $5,000 to double if it is invested at 4%, compounded

quarterly?P = $5000r = 4% or 0.04Compounded quarterly = 4 times per year, n = 4.The investment doubles, so A must be $10,000.

Substitute these values and solve for t.

Algebra II 25

Page 26: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

It takes more than17 years for themoney to double

in value.

Algebra II 26

Page 27: 1. Use a property of logarithms to evaluate log 3 9 2 2. Use log 5 ≈ 0.699 and log 6 ≈ 0.778 to approximate the value of log 150. 3. Expand ln 7 3 2x 4.

27Algebra II

Solve each equation. Be sure to check solutions!!

1.32x = 27x+2

2.5e3x + 2 = 17

3.log4(5x – 11) = log4(3 – 2x)

4.-3 ln = 4


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