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### Basic Logarithmic Functions

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1. Without a calculator give the exact value of each of the following logarithms.
1. $${\log _2}16$$
2. $${\log _4}16$$
3. $${\log _5}625$$
1. $$\displaystyle {\log _9}\frac{1}{{531441}}$$
2. $$\displaystyle {\log _{\frac{1}{6}}}36$$
3. $$\displaystyle {\log _{\frac{3}{2}}}\frac{{27}}{8}$$
Show Solution

To do these without a calculator you need to remember the following.

$y = {\log _b}x\hspace{0.25in}\hspace{0.25in}{\mbox{is equivalent to }}\hspace{0.25in}\hspace{0.25in}x = {b^y}$

Where, $$b$$, is called the base is any number such that $$b > 0$$ and $$b \ne 1$$. The first is usually called logarithmic form and the second is usually called exponential form. The logarithmic form is read “$$y$$ equals log base $$b$$ of $$x$$”.

So, if you convert the logarithms to exponential form it’s usually fairly easy to compute these kinds of logarithms.

(a) $${\log _2}16 = 4\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{2^4} = 16$$

(b) $${\log _4}16 = 2\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{4^2} = 16$$

Note the difference between (a) and (b)! The base, $$b$$, that you use on the logarithm is VERY important! A different base will, in almost every case, yield a different answer. You should always pay attention to the base!

(c) $${\log _5}625 = 4\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{5^4} = 625$$

(d) $$\displaystyle {\log _9}\frac{1}{{531441}} = - 6\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{9^{ - 6}} = \frac{1}{{{9^6}}} = \frac{1}{{531441}}$$

(e) $$\displaystyle {\log _{\frac{1}{6}}}36 = - 2\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{\left( {\frac{1}{6}} \right)^{ - 2}} = {6^2} = 36$$

(f) $$\displaystyle {\log _{\frac{3}{2}}}\frac{{27}}{8} = 3\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{\left( {\frac{3}{2}} \right)^3} = \frac{{27}}{8}$$

2. Without a calculator give the exact value of each of the following logarithms.
1. $$\ln \sqrt{{\bf{e}}}$$
2. $$\log 1000$$
3. $${\log _{16}}16$$
1. $${\log _{23}}1$$
2. $${\log _2}\sqrt{{32}}$$
Show Solution

There are a couple of quick notational issues to deal with first.

\begin{align*}\ln x & = {\log _{\bf{e}}}x & \hspace{0.5in} & {\mbox{This log is called the natural logarithm}}\\ \log x & = {\log _{10}}x & \hspace{0.5in} & {\mbox{This log is called the common logarithm}}\end{align*}

The $$e$$ in the natural logarithm is the same $$e$$ used in Problem 2 above. The common logarithm and the natural logarithm are the logarithms are encountered more often than any other logarithm so the get used to the special notation and special names.

The work required to evaluate the logarithms in this set is the same as in problem in the previous problem.

(a) $$\ln \sqrt{{\bf{e}}} = \frac{1}{3}\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{{\bf{e}}^{\frac{1}{3}}} = \sqrt{{\bf{e}}}$$

(b) $$\log 1000 = 3\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{10^3} = 1000$$

(c) $${\log _{16}}16 = 1\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{16^1} = 16$$

(d) $${\log _{23}}1 = 0\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}{23^0} = 1$$

(e) $${\log _2}\sqrt{{32}} = \frac{5}{7}\hspace{0.25in}\hspace{0.25in}{\rm{because}}\hspace{0.25in}\hspace{0.25in}\sqrt{{32}} = {32^{\frac{1}{7}}} = {\left( {{2^5}} \right)^{\frac{1}{7}}} = {2^{\frac{5}{7}}}$$