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log b ( M N ) = log b ( M ) + log b ( N ) .

Let m = log b M and n = log b N . In exponential form, these equations are b m = M and b n = N . It follows that

log b ( M N ) = log b ( b m b n ) Substitute for  M  and  N . = log b ( b m + n ) Apply the product rule for exponents . = m + n Apply the inverse property of logs . = log b ( M ) + log b ( N ) Substitute for  m  and  n .

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider log b ( w x y z ) . Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

log b ( w x y z ) = log b w + log b x + log b y + log b z

The product rule for logarithms

The product rule for logarithms    can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

log b ( M N ) = log b ( M ) + log b ( N )  for  b > 0

Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.

  1. Factor the argument completely, expressing each whole number factor as a product of primes.
  2. Write the equivalent expression by summing the logarithms of each factor.

Using the product rule for logarithms

Expand log 3 ( 30 x ( 3 x + 4 ) ) .

We begin by factoring the argument completely, expressing 30 as a product of primes.

log 3 ( 30 x ( 3 x + 4 ) ) = log 3 ( 2 3 5 x ( 3 x + 4 ) )

Next we write the equivalent equation by summing the logarithms of each factor.

log 3 ( 30 x ( 3 x + 4 ) ) = log 3 ( 2 ) + log 3 ( 3 ) + log 3 ( 5 ) + log 3 ( x ) + log 3 ( 3 x + 4 )
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Expand log b ( 8 k ) .

log b 2 + log b 2 + log b 2 + log b k = 3 log b 2 + log b k

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Using the quotient rule for logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: x a b = x a b . The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

Given any real number x and positive real numbers M , N , and b , where b 1 , we will show

log b ( M N ) = log b ( M ) log b ( N ) .

Let m = log b M and n = log b N . In exponential form, these equations are b m = M and b n = N . It follows that

log b ( M N ) = log b ( b m b n ) Substitute for  M  and  N . = log b ( b m n ) Apply the quotient rule for exponents . = m n Apply the inverse property of logs . = log b ( M ) log b ( N ) Substitute for  m  and  n .

For example, to expand log ( 2 x 2 + 6 x 3 x + 9 ) , we must first express the quotient in lowest terms. Factoring and canceling we get,

log ( 2 x 2 + 6 x 3 x + 9 ) = log ( 2 x ( x + 3 ) 3 ( x + 3 ) ) Factor the numerator and denominator .                         = log ( 2 x 3 ) Cancel the common factors .

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

log ( 2 x 3 ) = log ( 2 x ) log ( 3 )               = log ( 2 ) + log ( x ) log ( 3 )

The quotient rule for logarithms

The quotient rule for logarithms    can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

log b ( M N ) = log b M log b N

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

  1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
  2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
  3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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