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Log b MN 2 = log b M + log b N 2

Pre-Calc Lesson 5-6 Laws of Logarithms Remember---Logs are ‘ inverses ’ of exponentials. Therefore all the rules of exponents will also work for logs. Laws of Logarithms:

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Log b MN 2 = log b M + log b N 2

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  1. Pre-Calc Lesson 5-6 • Laws of Logarithms • Remember---Logs are ‘inverses’ of exponentials. • Therefore all the rules of exponents will also work for • logs. • Laws of Logarithms: • If M and N are positive real numbers and ‘b’ is a positive • number other than 1, then: • 1. logb MN = logb M + logb N • 2. logbM = logb M – logbN • N • logb M = logb N iff M = N • logb Mk = k logb M, for any real number k • Example 1:Express logbMN2 in terms of logbM and logbN • 1st: Recognize that you are taking the ‘log’ of a product  (M)(N2) • So we can split that up as an addition of two separate logs!

  2. Logb MN2 = logbM + logbN2 (Now recognize that we have a power on the number in the 2nd log. = logbM + 2logbN ! Example 2 : Express logbM3 in terms of logbM and logbN N LogbM3 = logb (M3)1/2 = ½ (logb(M3)) N (N) (N) = ½ (logbM3 – logbN) = ½ (3logbM – logbN) = 3/2 logbM – ½ logbN Example 3:Simplify log 45 – 2 log 3 log 45 – 2 log 3 = log 45 – log 32 = log 45 – log 9 = log (45/9) = log 5

  3. Example 4: Express y in terms of x if ln y = 1/3 ln x + ln 4 ln y = 1/3 ln x + ln 4 ln y = ln x1/3 + ln 4 ln y = ln (x1/3)(4) So the only way ln y = ln 4x1/3 is if : y = 4x1/3 Example 5: Solve log2x + log2(x – 2) = 3 log2x(x - 2) = 3 (Go to exponential form: 23 = x(x – 2) 8 = x2 – 2x 0 = x2 - 2x - 8 0 = (x – 4)(x + 2) x = 4, x = - 2 Now the domain of all log statements is (0, ф)  x ≠ - 2 so x = 4 is the only solution!!!

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