6.6: Logarithmic Properties

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

logb1=0logbb=1logb1=0logbb=1

For example, log51=0log51=0 since 50=1.50=1. And log55=1log55=1 since 51=5.51=5.

Next, we have the inverse property.

logb(bx)=x  blogbx=x,x>0logb(bx)=x  blogbx=x,x>0

For example, to evaluate log(100),log(100), we can rewrite the logarithm as log10(102),log10(102), and then apply the inverse property logb(bx)=xlogb(bx)=x to get log10(102)=2.log10(102)=2.

To evaluate eln(7),eln(7), we can rewrite the logarithm as eloge7,eloge7, and then apply the inverse property blogbx=xblogbx=x to get eloge7=7.eloge7=7.

Finally, we have the one-to-one property.

logbM=logbNif and only ifM=NlogbM=logbNif and only ifM=N

We can use the one-to-one property to solve the equation log3(3x)=log3(2x+5)log3(3x)=log3(2x+5) for x.x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x:x:

3x=2x+5x=5Set the arguments equal.Subtract 2x.3x=2x+5Set the arguments equal.x=5Subtract 2x.

But what about the equation log3(3x)+log3(2x+5)=2?log3(3x)+log3(2x+5)=2? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.

Recall that we use the product rule of exponents to combine the product of powers by adding exponents: xaxb=xa+b.xaxb=xa+b. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.

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

logb(MN)=logb(M)+logb(N).logb(MN)=logb(M)+logb(N).

Let m=logbMm=logbM and n=logbN.n=logbN. In exponential form, these equations are bm=Mbm=M and bn=N.bn=N. It follows that

logb(MN)=logb(bmbn)=logb(bm+n)=m+n=logb(M)+logb(N)Substitute for Mand N.Apply the product rule for exponents.Apply the inverse property of logs.Substitute for mand n.logb(MN)=logb(bmbn)Substitute for Mand N.=logb(bm+n)Apply the product rule for exponents.=m+nApply the inverse property of logs.=logb(M)+logb(N)Substitute for mand 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 logb(wxyz).logb(wxyz). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

logb(wxyz)=logbw+logbx+logby+logbzlogb(wxyz)=logbw+logbx+logby+logbz

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

logb(MN)=logb(M)+logb(N)for b>0logb(MN)=logb(M)+logb(N)for b>0

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

Expand log3(30x(3x+4)).log3(30x(3x+4)).

Expand logb(8k).logb(8k).

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: xaxb=xa−b.xaxb=xa−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 xx and positive real numbers M,M, N,N, and b,b, where b≠1,b≠1, we will show

logb(MN)=logb(M)−logb(N).logb(MN)=logb(M)−logb(N).

Let m=logbMm=logbM and n=logbN.n=logbN. In exponential form, these equations are bm=Mbm=M and bn=N.bn=N. It follows that

logb(MN)=logb(bmbn)=logb(bm−n)=m−n=logb(M)−logb(N)Substitute for Mand N.Apply the quotient rule for exponents.Apply the inverse property of logs.Substitute for mand n.logb(MN)=logb(bmbn)Substitute for Mand N.=logb(bm−n)Apply the quotient rule for exponents.=m−nApply the inverse property of logs.=logb(M)−logb(N)Substitute for mand n.

For example, to expand log(2x2+6x3x+9),log(2x2+6x3x+9), we must first express the quotient in lowest terms. Factoring and canceling we get,

log(2x2+6x3x+9)=log(2x(x+3)3(x+3))                      =log(2x3)Factor the numerator and denominator.Cancel the common factors.log(2x2+6x3x+9)=log(2x(x+3)3(x+3))Factor the numerator and denominator.                      =log(2x3)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(2x3)=log(2x)−log(3)            =log(2)+log(x)−log(3)log(2x3)=log(2x)−log(3)            =log(2)+log(x)−log(3)

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

logb(MN)=logbM−logbNlogb(MN)=logbM−logbN

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

Expand log2(15x(x−1)(3x+4)(2−x)).log2(15x(x−1)(3x+4)(2−x)).

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for x=−43x=−43 and x=2.x=2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that x>0,x>0, x>1,x>1, x>−43,x>−43, and x1 and n>1.n>1.

42. 

Does log81(2401)=log3(7)?log81(2401)=log3(7)? Verify the claim algebraically.