### 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#.

Simplify \[

\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{6 \pi}}\right)

\]to an expression without logarithms.

\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{6 \pi}}\right)

\]to an expression without logarithms.

Solution #\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{6 \pi}}\right)=# #{{17}\over{2}}-6 \pi#

This can be seen from the following calculation:

\[\begin{array}{rcl}

\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{6 \pi}}\right) &=&\log_{3}({3}^{\frac{1}{2}})+4\cdot\log_{3}({3}^{2}) -6 \pi\cdot\log_{3}(3) \\&=& \frac{1}{2} +4\cdot 2 -6 \pi\\ &=& {{17}\over{2}}-6 \pi

\end{array}

\]

This can be seen from the following calculation:

\[\begin{array}{rcl}

\log_{3}\left(\frac{\sqrt{3}\cdot{9}^{4}}{{3}^{6 \pi}}\right) &=&\log_{3}({3}^{\frac{1}{2}})+4\cdot\log_{3}({3}^{2}) -6 \pi\cdot\log_{3}(3) \\&=& \frac{1}{2} +4\cdot 2 -6 \pi\\ &=& {{17}\over{2}}-6 \pi

\end{array}

\]

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