Logarithms and Equation Solving Chapter 2 Exponents and Logarithms 2.7A Logarithms and Equation Solving MATHPOWERTM 12, WESTERN EDITION 2.7A.1

Solving Log Equations 1. log272 = log2x + log212 2. 2x = 8 log272 - log212 = log2x 2. 2x = 8 log 2x = log 8 xlog2 = log 8 x = 3 x = 6 3. x = 3.79 xlog7 = 2log40 2.7A.2

4. log7(2x + 2) - log7(x - 1) = log7(x + 1) Solving Log Equations 4. log7(2x + 2) - log7(x - 1) = log7(x + 1) 2x + 2 = (x- 1)(x + 1) 2x + 2 = x2 - 1 0 = x2 - 2x - 3 0 = (x - 3)(x + 1) x - 3 = 0 or x + 1 = 0 x = 3 x = -1 Therefore, x = 3. Check: log7(2x + 2) - log7(x - 1) = log7(x + 1) log7(2x + 2) - log7(x - 1) = log7(x + 1) log7(2(3) + 2) - log7(3 - 1) = log7(3 + 1) log7(2(-1) + 2) - log7(-1 - 1) = log7(-1 + 1) log70 - log7(-2) = log7(0) Negative logarithms and logs of 0 are undefined. log74 = log74 2.7A.3

Solving Log Equations 5. log7(x + 1) + log7(x - 5) = 1 log7[(x + 1)(x - 5)] = log77 (x + 1)(x - 5) = 7 x2 - 4x - 5 = 7 x2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 x - 6 = 0 or x + 2 = 0 x = 6 x = -2 6. log4(4x) - log4(x - 2) = 3 64(x - 2) = 4x 64x - 128 = 4x 60x = 128 x = 2.13 x = -2 is extraneous. Therefore, x = 6. 2.7A.4

Solve: log5(x - 6) = 1 - log5(x - 2) Solving Log Equations Express 12 as a power of 2: Solve: log5(x - 6) = 1 - log5(x - 2) log5(x - 6) + log5(x - 2) = 1 log5(x - 6)(x - 2) = 1 log5(x - 6)(x - 2) = log551 (x - 6)(x - 2) = 5 x2 - 8x + 12 = 5 x2 - 8x + 7 = 0 (x - 7)(x - 1) = 0 x = 7 or x = 1 xlog2 = log12 x = 3.58 23.58 = 12 Since x > 6, the value of x = 1 is extraneous. Therefore, the solution is x = 7. 2.7A.5

Solving Log Equations 7. 3x = 2x + 1 log(3x) = log(2x + 1) x log 3 = x log 2 + 1 log 2 x log 3 - x log 2 = log 2 x(log 3 - log 2) = log 2 8. 23x - 1 = 32x - 1 (3x - 1) log 2 = (2x - 1) log 3 3x log 2 -1 log2 = 2x log3 - log3 3xlog2 - 2xlog3 = log2 - log3 x(3log2 - 2log3) = log2 - log3 x = 3.44 x = 1.71 2.7A.6

Applications of Logarithms 1. Carbon 14 has a half-life of 5760 years. Find the age of a specimen with 24% C-14 relative to living matter. t = 11 859.23 Therefore, the specimen is 11 859.23 years old. 2.7A.7

Applications of Logarithms Find the time period required for $7000 invested at 10%/a compounded semi-annually to grow to $10 000. A(t) = P(1 + i) 2n 10 000 = 7000(1.05)2n log10 - log7 = 2nlog1.05 7.31 = 2n 3.66 = n It would take 3.66 years for the investment to grow to $10 000. 2.7A.8

Applications of Logarithms 3. The value of an investment is given by f(x) = 237.50(1.052)x, where x is the number of 6-month periods. Find the number of complete periods until the investment is worth at least $600. f(x) = 237.50(1.052)x 600 = 237.50(1.052)x 2.5263 = (1.052)x log 2.5263 = x log 1.052 x = 18.28 Therefore, after 19 periods the investment would be worth at least $600. 2.7A.9

36 = t Applications of Logarithms 4. Cell population doubles every 3 h. How long would it take 4 cells to reach a count of 16 384? It would take 36 h to reach 16 384 cells. 36 = t 2.7A.10

Applications of Logarithms 5. For every metre below the water surface, light intensity is reduced by 5%. At what depth is light intensity 40% of that at the surface? Id = Io(1 - 0.05)d 40 = 100(0.95)d 0.4 = 0.95d log 0.4 = dlog0.95 d = 17.86 Therefore, at a depth of 17.86 m the light intensity would be 40%. 2.7A.11

More Applications - Comparing Intensities of Sound For any intensity, I, the decibel level, dB, is defined as follows: where Io is the intensity of a barely audible sound 6. The sound at a rock concert is 106 dB. During the break, the sound is 76 dB. How many times as loud is it when the band is playing? Louder Softer Comparison Thus, it would be 1000 times as loud. I = 107.6 Io I = 1010.6 Io 2.7A.12

More Applications - The Richter Scale I = Io(10)m where m is the measure on the scale 7. Compare the intensities of the Japan earthquake of 1933, which measured 8.9 on the Richter Scale, to the earthquake of Turkey in 1966, which measured 6.9 on the scale. Therefore, the earthquake in Japan is 100 times as intense as the one in Turkey. 2.7A.13

More Applications - The Richter Scale 8. The magnitude of earthquakes is given by where I is the quake intensity and Io is the reference intensity. How many times as intense is a quake of 8.1 compared to a quake with a magnitude of 6.4? Comparison Therefore, a quake of 8.1 is 50.1 times as great. 2.7A.14

More Applications - The Richter Scale 9. Earthquake intensity is given by I = Io (10)m, where m is the magnitude and Io is the relative intensity. A quake of magnitude 7.9 is 120 times as intense as a tremor. What is the magnitude of the tremor? Iq = Io (10)7.9 It = Io (10)m log 120 = (7.9 - m) log 10 log 120 = (7.9 - m) The magnitude of the tremor is 5.8. m = 7.9 - log 120 m = 5.8 2.7A.15

Assignment Suggested Questions: Pages 113 and 114 A 1-13 odd, 21-25, 40 a-e B 26-33, 40 f-h, 42-44 Page 99 46 2.7A.16

Logarithmic Identities Chapter 2 Exponents and Logarithms 2.7B Logarithmic Identities MATHPOWERTM 12, WESTERN EDITION 2.7B.17

Some equations are solved for all values of the variable for Proving Identities Some equations are solved for all values of the variable for Which both sides of the equation are defined. These equations are called identities. = logax -1 = -1logax -logax = -logax L.S. = R.S. Therefore, the identity is proved. L.S. = R.S. 2.7B.18

Proving Identities Prove: = Conclusion: This suggests that L.S. = R.S. 2.7B.19

Proving Identities Verify: L.S. = R.S. 2.7B.20

Assignment Suggested Questions: Page 106 44-46 Page 113 33-38 2.7B.21