Download Math 142–Rodriguez Lehmann – 5.6

Math 142–Rodriguez
Lehmann – 5.6
More Properties of Logarithms
In section 5.4 we learned how to solve logarithmic equations that only had one logarithm in
the equation.
3log27(x)+7=8
logb(80)=3
In this section we learn how to solve different logarithmic equations from those above. In
order to do this we first need to discuss some other properties of logarithms.
Product Property for Logarithms
For x>0, y>0, b>0, and b≠1, logb(x) + logb(y) = logb(xy)
Proof: Done in book; I will do on board; sit back and watch.
Examples: Simplify. Write your result as a single logarithm with a coefficient of 1.
1. logb(3x) + logb(y)
3. 4logb(3x) + 3logb(2x4)
2. 5logb(x) + logb(2x)
Quotient Property for Logarithms
x
For x>0, y>0, b>0, and b≠1, logb(x) – logb(y) = logb( y )
Proof: not done in book or in class; similar to previous proof
Examples: same instructions as before
1. logb(27x) – logb(3)
2. logb(4x) – 5logb(x)
3. 5logb(x) – logb(x)
Solving Logarithmic Equations
Steps:
1. Use the Product or Quotient Property to rewrite the logarithms as a single logarithm.
2. Now that the equation looks like: logb(#)=# rewrite in exponential form.
3. Solve the resulting equation.
Solve. Round any solutions to the fourth decimal place.
Yours:
Lehmann – 5.6
Page 2 of 3
Change–of–Base Property
This property has two uses: use our calculators to find the logarithm of any number and any
base; and could use to solve yet another type of logarithmic equation (if time I will show you
one example).
For a>0, b>0, a≠1, b≠1, and x>0, log b (x) =
log a (x)
.
log a (b)
Proof: done in book; will do on board if we have time
Example: Evaluate your result to the fourth decimal place.
1. log5(63)
Lehmann – 5.6
2. log8(101)
3. log2(35.4)
Page 3 of 3