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### 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.