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 \(0 \lt b \lt 1\) then it increases as we move left on the \(x\)-axis. (The function \(f(x) = b^x\) is not very exciting if \(b = 1\text{.}\)) In either case, that means for any positive real number \(y\) there exists a unique real number \(x\) such that \(b^x = y\text{.}\) This number is the base-\(b\) logarithm of \(y\), and denoted by \(\log_b(y)\text{.}\) That is, if \(b > 0\) and \(b \neq 1\text{,}\) then \(\log_b(y)\) is the power to which \(b\) must be raised in order to get the positive number \(y\text{.}\)
It is important to notice that \(\log_b(y)\) is only defined for positive numbers \(y\text{.}\) Logarithms are exponents. By definition \(b^{\log_b(y)} = y\text{.}\) Since \(b\) is positive, so is any power of \(b\text{.}\) Thus \(\log_b(y)\) 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 \(b \neq 1\text{.}\)
- \(\log_b{1} = 0\) because \(b^0 = 1\text{.}\)
- \(\log_b{b} = 1\) because \(b^1 = b\text{.}\)
- \(b^x = y\) if and only if \(\log_b(y) = x\text{.}\)
- \(\log_b(xy) = \log_b(x) + \log_b(y)\) because \(b^{\log_b(x) + \log_b(y)} = b^{log_b(x)} b^{\log_b(y)} = xy\text{.}\)
- \(\log_b(a^x) = x \log_b(a)\) because \(b^{x \log_b(a)} = \left(b^{\log_b(a)}\right)^x = a^x\text{.}\)
- \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\) because \(b^{ \log_b(x) - \log_b(y)} = b^{log_b(x)} b^{-\log_b(y)} = b^{log_b(x)} \cdot \frac{1}{b^{\log_b(y)}} = \frac{x}{y}\text{.}\)
When the base \(b\) is omitted, as in \(\log(100)\text{,}\) 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 \(log_e(x)\text{.}\)
Evaluate each expression.
- \(\displaystyle \log_2{(2)} = 1\)
- \(\displaystyle \log{(100)} = \log{(10^2)} = \log_{10}{(10^2)} = 2\)
- \(\displaystyle \log_{6}{(12)} + \log_{6}{(18)} = \log_{6}{(12 \cdot 18)} = \log_{6}{(216)} = \log_{6}{(6^3)} = 3\)
- \(\displaystyle \log_{5}{(2500)} - \log_{5}{(4)} = \log_{5}{(\frac{2500}{4})} = \log_{5}{(625)} = \log_{5}{(5^4)} = 4\)
- \(\displaystyle \log(5b) + \log(2c^2) = \log(10bc^2) = \log(10) + \log(bc^2) = 1 + \log(bc^2)\)
Example 4.4.2. Solving Equations with Logarithms.
Solve \(\log_8(x) + \log_8(x - 12) = 2\text{.}\)
Solution.
We have \(2 = \log_8(x) + \log_8(x - 12) = \log_8(x(x - 12))\text{,}\) so that \(8^2 = x(x - 12)\text{,}\) by the definition of logarithms. Thus \(x^2 - 12 - 64 = 0\text{.}\) By factoring (or using the quadratic formula first for help), this is the same as \((x-16)(x+4) = 0\text{.}\) Therefore \(x = 16\) or \(x = -4\text{.}\) But \(x = -4\) is not a solution, as \(\log_8(-4)\) is undefined. Hence the solution is \(x = 16\text{.}\)
For a positive real number \(b \neq 1\text{,}\) the logarithm function with base \(b\) is the function \(f(x) = \log_b(x)\text{.}\) Its domain is the set of positive real numbers. Its range is the set of all real numbers.
Notice that the functions \(f(x) = b^x\) and \(g(y) = \log_b(y)\) are inverses by definition: \(b^x = y\) if and only if \(\log_b(y) = x\text{.}\) Thus, if \(b \neq 1\text{,}\) the logarithm function with base \(b\) is the inverse of the exponential function with base \(b\text{.}\) (The function \(f(x) = 1^x\) does not have an inverse because, for example \(1^2 = 1^3 = 1\text{.}\))
Finally, we observe that logarithm functions with different bases are just multiples of each other. We know \(b^{\log_b(x)} = x\text{.}\) Therefore, \(\log_a\left(b^{\log_b(x)}\right) = \log_a(x)\text{.}\) From one of the properties of logarithms, we know that \(\log_a\left(b^{\log_b(x)}\right) =\log_b(x) \log_a(b)\text{,}\) and so this is the same as \(\log_b(x) \log_a(b) = \log_a(x)\text{.}\) Since \(\log_a(b)\) is a number, this says that \(\log_a(x)\) is a multiple of \(\log_b(x)\text{.}\)
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 \(f(x) = 5\log(x+2) + 3\text{.}\)
Solution.
This graph has the same basic shape as the graph of \(y = \log(x)\text{.}\) 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.
\(\log_{6}{(12)} + \log_{6}{(18)}\)
Answer.
\(3\)
2.
\(\log_8(x) + \log_8(x - 12)\)
Answer.
\(\log_{8}{(x^2-12x)}\)
3.
\(\log_{5}{(2500)} - \log_{5}{(4)}\)
Answer.
\(4\)
4.
\(\log(5b) + \log(2c^2)\)
Answer.
\(\log(10bc^2)\)
5.
\(\log_{25}{(7)} + \log_5{(3)}\)
Answer.
\(\frac{2\log{(3)}+\log{(7)}}{2\log{(5)}}\)
6.
Sketch the graph of \(h(x)=1-5\log{(1-\frac{x}{2})}\text{.}\)
