Skip to main content

Precalculus Review Materials

Section 4.4 Logarithms

From our previous work, we know that if b>0 then the range of the exponential function f(x)=b(x) is the positive real numbers. If b>0 then the graph of f increases as we move right on the x axis, and if 0<b<1 then it increases as we move left on the x-axis. (The function f(x)=bx is not very exciting if b=1.) In either case, that means for any positive real number y there exists a unique real number x such that bx=y. This number is the base-b logarithm of y, and denoted by logb(y). That is, if b>0 and b1, then logb(y) is the power to which b must be raised in order to get the positive number y.
It is important to notice that logb(y) is only defined for positive numbers y. Logarithms are exponents. By definition blogb(y)=y. Since b is positive, so is any power of b. Thus logb(y) is undefined when y is negative because no power of a positive number can give us a negative number.
Logarithms have properties that follow immediately from the fact that they are exponents.
Properties of Logarithms. Suppose b>0 and b1.
  • logb1=0 because b0=1.
  • logbb=1 because b1=b.
  • bx=y if and only if logb(y)=x.
  • logb(xy)=logb(x)+logb(y) because blogb(x)+logb(y)=blogb(x)blogb(y)=xy.
  • logb(ax)=xlogb(a) because bxlogb(a)=(blogb(a))x=ax.
  • logb(xy)=logb(x)logb(y) because blogb(x)logb(y)=blogb(x)blogb(y)=blogb(x)1blogb(y)=xy.
When the base b is omitted, as in log(100), it is assumed to be 10. Logarithms to base 10 are called common logarithms. Logarithms to base e are called natural logarithms, and denoted by ln(x) rather than loge(x).

Example 4.4.1. Evaluating Logarithms.

Evaluate each expression.
  • log2(2)=1
  • log(100)=log(102)=log10(102)=2
  • log6(12)+log6(18)=log6(1218)=log6(216)=log6(63)=3
  • log5(2500)log5(4)=log5(25004)=log5(625)=log5(54)=4
  • log(5b)+log(2c2)=log(10bc2)=log(10)+log(bc2)=1+log(bc2)

Example 4.4.2. Solving Equations with Logarithms.

Solve log8(x)+log8(x12)=2.
Solution.
We have 2=log8(x)+log8(x12)=log8(x(x12)), so that 82=x(x12), by the definition of logarithms. Thus x212x64=0. By factoring (or using the quadratic formula first for help), this is the same as (x16)(x+4)=0. Therefore x=16 or x=4. But x=4 is not a solution, as log8(4) is undefined. Hence the solution is x=16.
For a positive real number b1, the logarithm function with base b is the function f(x)=logb(x). Its domain is the set of positive real numbers. Its range is the set of all real numbers.
Notice that the functions f(x)=bx and g(y)=logb(y) are inverses by definition: bx=y if and only if logb(y)=x. Thus, if b1, the logarithm function with base b is the inverse of the exponential function with base b. (The function f(x)=1x does not have an inverse because, for example 12=13=1.)
Finally, we observe that logarithm functions with different bases are just multiples of each other. We know blogb(x)=x. Therefore, loga(blogb(x))=loga(x). From one of the properties of logarithms, we know that loga(blogb(x))=logb(x)loga(b), and so this is the same as logb(x)loga(b)=loga(x). Since loga(b) is a number, this says that loga(x) is a multiple of logb(x).
The graphs of f(x)=log(x) and g(x)=ln(x) are shown below.
The same principles as before of shifting graphs or stretching them vertically or horizontally apply.

Example 4.4.3. Graphing Logarithmic Functions.

Sketch the graph of f(x)=5log(x+2)+3.
Solution.
This graph has the same basic shape as the graph of y=log(x). It is shifted upwards by 3 and left by 2. Also, it is stretched in the vertical direction by a factor of 5.
After plotting a few well chosen points and sketching a curve of the correct shape through them, one arrives at the graph below.

Exercises Practice Problems

In questions 1 to 5, use the properties of logarithms to to find an equivalent, arguably simpler, expression.

1.

log6(12)+log6(18)
Answer.
3

2.

log8(x)+log8(x12)
Answer.
log8(x212x)

3.

log5(2500)log5(4)
Answer.
4

4.

log(5b)+log(2c2)
Answer.
log(10bc2)

5.

log25(7)+log5(3)
Answer.
2log(3)+log(7)2log(5)

6.

Sketch the graph of h(x)=15log(1x2).
Answer.