Section 3.3 : Logarithm Properties

Complete the following formulas. Show All Solutions Hide All Solutions

1. ${\log _b}b =$
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   $${\log _b}b = 1\hspace{0.5in}{\rm{because}}\hspace{0.5in}{b^1} = b$$
2. ${\log _b}1 =$
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   $${\log _b}1 = 0\hspace{0.5in}{\rm{because}}\hspace{0.5in}{b^0} = 1$$
3. ${\log _b}{b^x} =$
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   $${\log _b}{b^x} = x$$
4. ${b^{{{\log }_b}x}} =$
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   $${b^{{{\log }_b}x}} = x$$
5. ${\log _b}xy =$
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   $${\log _b}xy = {\log _b}x + {\log _b}y$$

   THERE IS NO SUCH PROPERTY FOR SUMS OR DIFFERENCES!!!!!

   $$\begin{align*}{\log _b}\left( {x + y} \right) & \ne {\log _b}x + {\log _b}y\\ {\log _b}\left( {x - y} \right) & \ne {\log _b}x - {\log _b}y\end{align*}$$
6. ${\log _b}\left( {\frac{x}{y}} \right) =$
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   $${\log _b}\left( {\frac{x}{y}} \right) = {\log _b}x - {\log _b}y$$

   THERE IS NO SUCH PROPERTY FOR SUMS OR DIFFERENCES!!!!!

   $$\begin{align*}{\log _b}\left( {x + y} \right) &\ne {\log _b}x + {\log _b}y\\ {\log _b}\left( {x - y} \right) &\ne {\log _b}x - {\log _b}y\end{align*}$$
7. ${\log _b}\left( {{x^r}} \right) =$
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   $${\log _b}\left( {{x^r}} \right) = r{\log _b}x$$

   Note in this case the exponent needs to be on the WHOLE argument of the logarithm. For instance,

   $${\log _b}{\left( {x + y} \right)^2} = 2{\log _b}\left( {x + y} \right)$$

   However,

   $${\log _b}\left( {{x^2} + {y^2}} \right) \ne 2{\log _b}\left( {x + y} \right)$$
8. Write down the change of base formula for logarithms.
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   The change of base formula for logarithms is,

   $${\log _b}x = \frac{{{{\log }_a}x}}{{{{\log }_a}b}}$$

   This is the most general change of base formula and will convert from base $b$ to base $a$. However, the usual reason for using the change of base formula is so you can compute the value of a logarithm that is in a base that you can’t easily compute. Using the change of base formula means that you can write the logarithm in terms of a logarithm that you can compute. The two most common change of base formulas are

   $${\log _b}x = \frac{{\ln x}}{{\ln b}}\hspace{0.5in}{\rm{and}}\hspace{0.5in}{\rm{lo}}{{\rm{g}}_{\rm{b}}}x = \frac{{\log x}}{{\log b}}$$

   In fact, often you will see one or the other listed as THE change of base formula!

   In the problems in the Basic Logarithm Functions section you computed the value of a few logarithms, but you could do these without the change of base formula because all the arguments could be written in terms of the base to a power. For instance,

   $${\log _7}49 = 2\hspace{0.5in}{\rm{because}}\hspace{0.5in}{7^2} = 49$$

   However, this only works because 49 can be written as a power of 7! We would need the change of base formula to compute ${\log _7}50$.

   $${\log _7}50 = \frac{{\ln 50}}{{\ln 7}} = \frac{{3.91202300543}}{{1.94591014906}} = 2.0103821378$$

   OR

   $${\log _7}50 = \frac{{\log 50}}{{\log 7}} = \frac{{1.69897000434}}{{0.845098040014}} = 2.0103821378$$

   So, it doesn’t matter which we use, you will get the same answer regardless.

   Note as well that we could use the change of base formula on ${\log _7}49$ if we wanted to as well.

   $${\log _7}49 = \frac{{\ln 49}}{{\ln 7}} = \frac{{3.89182029811}}{{1.94591014906}} = 2$$

   This is a lot of work however, and is probably not the best way to deal with this.

9. What is the domain of a logarithm?
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   The domain $y = {\log _b}x$ is $x > 0$. In other words, you can’t plug in zero or a negative number into a logarithm. This makes sense if you remember that $b > 0$ and write the logarithm in exponential form.

   $$y = {\log _b}x\hspace{0.25in} \Rightarrow \hspace{0.25in}{b^y} = x$$

   Since $b > 0$ there is no way for $x$ to be either zero or negative. Therefore, you can’t plug a negative number or zero into a logarithm!

10. Sketch the graph of $f\left( x \right) = \ln \left( x \right)$ and $g\left( x \right) = \log \left( x \right)$.
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    Not much to this other than to use a calculator to evaluate these at a few points and then make the sketch. Here is the sketch.

    

    From this graph we can see the following behaviors of each graph.

    $$\begin{align*} & \ln \left( x \right) \to \infty {\mbox{ as }}x \to \infty & \hspace{0.5in} & {\rm{and}} & \hspace{0.5in} & \ln \left( x \right) \to - \infty {\mbox{ as }}x \to 0\,\,\left( {x > 0} \right)\\ & \log \left( x \right) \to \infty {\mbox{ as }}x \to \infty & \hspace{0.5in} & {\rm{and}} & \hspace{0.5in} & \log \left( x \right) \to - \infty {\mbox{ as }}x \to 0\,\,\left( {x > 0} \right)\end{align*}$$

    Remember that we require $x > 0$ in each logarithm.