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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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Solving Exponential and Logarithmic Equations Algebra II
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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” 3 Algebra II
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a x = a y If x = y Log b x = log b y if x = y Examples: If 2 x = 2 3 then x = 3 If ln x = 3 then e lnx = e 3, so x = e 3 If log 3 x = log 3 (5), then x = 5 If e x = 7 then ln e x = ln7, so x = ln7 4 Algebra II
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10 Algebra II
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11 Algebra II
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12 Algebra II No Solution (Extraneous solution)
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13 Algebra II
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14 Algebra II
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15 Algebra II
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16 Algebra II
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17 Algebra II Check both solutions!
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18 Algebra II
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19 Algebra II
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20 Algebra II
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21 Algebra II
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22 Algebra II
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ln x = x 2 – 2 Log 3 x = x 2 – 2 23 Algebra II
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The population size y of a community of lemmings varies according to the relationship y = y 0 e 0.15t. In this formula, t is time in months and y 0 is the initial population at the time 0. Estimate the population after 8 months if there were originally 2000 lemmings. y = y 0 e 0.15t y = 2000e 0.15(8) y = 2000e 1.2 y ≈ 6640.2339 In 8 months, the population will be approximately 6640 lemmings. Algebra II 24
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How long does it take an investment of $5,000 to double if it is invested at 4%, compounded quarterly? P = $5000 r = 4% or 0.04 Compounded 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
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It takes more than 17 years for the money to double in value. Algebra II 26
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27 Algebra II Solve each equation. Be sure to check solutions!! 1.3 2x = 27 x+2 2.5e 3x + 2 = 17 3.log 4 (5x – 11) = log 4 (3 – 2x) 4.-3 ln = 4
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