Operations for functions: Exponential and logarithmic functions
Properties of logarithms
Properties of the logarithm
Let #a# and #b# be positive numbers distinct from #1#. Then, for positive real numbers #x#, #y#, the following rules hold:
| 1 | \(\log_a(x\cdot y) = \log_a (x) + \log_a (y)\) |
| 2 | \(\log_a\frac{x}{y} = \log_a (x) - \log_a (y)\) |
| 3 | \(\log_a\left(x^p\right) =p\cdot \log_a(x)\) |
| 4 | \(\log_a(a)=1\) |
| 5 | \(\log_a(1)=0\) |
| 6 | \(\log_a(x) = \frac{\log_b(x)}{\log_b (a)}\) |
| 7 | if #a\lt b# and #x\gt 0#, then #\log_a(x)\lt\log_b(x)# |
| 8 | if #a\gt 1# and #x\lt y#, then #\log_a(x)\lt\log_ a(y)# |
In particular, #\log_a(x)# is increasing if #a\gt1# and decreasing if #a\lt 1#.
These rules are direct consequences of the known calculation rules for arithmetic and the properties of exponents.
Here are some examples:
| 1 | \(\log_2(8\cdot 16) = \log_2 (8) + \log_2 (16)=3+4=7\) |
| 2 | \(\log_2\left (\frac{8}{16} \right) = \log_2 (8) - \log_2 (16)=3-4=1\) |
| 3 | \(\log_2(8)^3 = 3\cdot \log_2 (8) =3.3=9\) |
| 4 | \(\log_2(2)=1\) |
| 5 | \(\log_3(1)=0\) |
| 6 | \(\log_2(8) =\frac{\log_{10}(8)}{\log_{10}(2)}=3\) |
Most scientific calculators have the predefined symbols of \(\log\) and \(\ln\). The symbol \(\log\) in the calculators is usually the logarithm to the base 10, \(\log_{10}\), and the symbol \(\ln\) is the logarithm to the base \(e\), that is, \(\log_e\).
Rule 6 makes it possible to express a logarithm with base #a# in logarithms with base #b#.
#\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{7 \pi}}\right)=# #{{17}\over{2}}-7 \pi#
This can be seen from the following calculation:
\[\begin{array}{rcl}
\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{7 \pi}}\right) &=&\log_{3}({3}^{\frac{1}{2}})+4\cdot\log_{3}({3}^{2}) -7 \pi\cdot\log_{3}(3) \\&=& \frac{1}{2} +4\cdot 2 -7 \pi\\ &=& {{17}\over{2}}-7 \pi
\end{array}
\]
This can be seen from the following calculation:
\[\begin{array}{rcl}
\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{7 \pi}}\right) &=&\log_{3}({3}^{\frac{1}{2}})+4\cdot\log_{3}({3}^{2}) -7 \pi\cdot\log_{3}(3) \\&=& \frac{1}{2} +4\cdot 2 -7 \pi\\ &=& {{17}\over{2}}-7 \pi
\end{array}
\]
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