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

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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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Presentation Transcript

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!

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

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!!!