3.3 Logarithms

Example 1

1. \displaystyle \lg 100000 = 5\quad because \displaystyle 10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}} = 100\,000.
2. \displaystyle \lg 0\textrm{.}0001 = -4\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} = 0\textrm{.}0001.
3. \displaystyle \lg \sqrt{10} = \frac{1}{2}\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}.
4. \displaystyle \lg 1 = 0\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
5. \displaystyle \lg 10^{78} = 78\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}} = 10^{78}.
6. \displaystyle \lg 50 \approx 1\textrm{.}699\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\textrm{.}699\,}} \approx 50.
7. \displaystyle \lg (-10) does not exist because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,a\vphantom{b,}\,}} can never be -10 regardless of how \displaystyle a is chosen.

Example 2

1. \displaystyle 10^{\textstyle\,\lg 100} = 100
2. \displaystyle 10^{\textstyle\,\lg a} = a
3. \displaystyle 10^{\textstyle\,\lg 50} = 50

Example 3

1. \displaystyle \log_{\,2} 8 = 3\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 8.
2. \displaystyle \log_{\,2} 2 = 1\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = 2.
3. \displaystyle \log_{\,2} 1024 = 10\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,10\vphantom{,}\,}} = 1024.
4. \displaystyle \log_{\,2}\frac{1}{4} = -2\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{2^2} = \frac{1}{4}.

Example 4

1. \displaystyle \log_{\,3} 9 = 2\quad because \displaystyle 3^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2\vphantom{,}\,}} = 9.
2. \displaystyle \log_{\,5} 125 = 3\quad because \displaystyle 5^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 125.
3. \displaystyle \log_{\,4} \frac{1}{16} = -2\quad because \displaystyle 4^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{4^2} = \frac{1}{16}.
4. \displaystyle \log_{\,b} \frac{1}{\sqrt{b}} = -\frac{1}{2}\quad as \displaystyle b^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-1/2\,}} = \frac{1}{b^{1/2}} = \frac{1}{\sqrt{b}} (if \displaystyle b>0 and \displaystyle b\not=1).

Example 5

1. \displaystyle \ln 10 \approx 2{,}3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{,}3\,}} \approx 10.
2. \displaystyle \ln e = 1\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e.
3. \displaystyle \ln\frac{1}{e^3} = -3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-3\vphantom{,}\,}} = \frac{1}{e^3}.
4. \displaystyle \ln 1 = 0\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
5. If \displaystyle y= e^{\,a} then \displaystyle a = \ln y.
6. \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5
7. \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x

Example 6

Calculate \displaystyle \,35\cdot 54.

If we know that \displaystyle 35 \approx 10^{\,1\textrm{.}5441} and \displaystyle 54 \approx 10^{\,1\textrm{.}7324} (i.e. \displaystyle \lg 35 \approx 1\textrm{.}5441 and \displaystyle \lg 54 \approx 1\textrm{.}7324) then we can calculate that

 35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324}
             = 10^{\,1\textrm{.}5441 + 1\textrm{.}7324}
             = 10^{\,3\textrm{.}2765}

and we then know that \displaystyle 10^{\,3\textrm{.}2765} \approx 1890 (i.e. \displaystyle \lg 1890 \approx 3\textrm{.}2765) thus we have managed to calculate the product

and this just by adding together exponents \displaystyle 1\textrm{.}5441 and \displaystyle 1\textrm{.}7324.

Example 7

1. \displaystyle \lg 4 + \lg 7 = \lg(4 \cdot 7) = \lg 28
2. \displaystyle \lg 6 - \lg 3 = \lg\frac{6}{3} = \lg 2
3. \displaystyle 2 \cdot \lg 5 = \lg 5^2 = \lg 25
4. \displaystyle \lg 200 = \lg(2 \cdot 100) = \lg 2 + \lg 100 = \lg 2 + 2

Example 8

1. \displaystyle \lg 9 + \lg 1000 - \lg 3 + \lg 0{,}001 = \lg 9 + 3 - \lg 3 - 3 = \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3
2. \displaystyle \ln\frac{1}{e} + \ln \sqrt{e} = \ln\left(\frac{1}{e} \cdot \sqrt{e}\,\right) = \ln\left( \frac{1}{(\sqrt{e}\,)^2} \cdot \sqrt{e}\,\right) = \ln\frac{1}{\sqrt{e}}
   \displaystyle \phantom{\ln\frac{1}{e} + \ln \sqrt{e}}{} = \ln e^{-1/2} = -\frac{1}{2} \cdot \ln e =-\frac{1}{2} \cdot 1 = -\frac{1}{2}\vphantom{\biggl(}
3. \displaystyle \log_2 36 - \frac{1}{2} \log_2 81 = \log_2 (6 \cdot 6) - \frac{1}{2} \log_2 (9 \cdot 9)
   \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = \log_2 (2\cdot 2 \cdot 3 \cdot 3) - \frac{1}{2} \log_2 (3 \cdot 3 \cdot 3 \cdot 3)
   \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = \log_2 (2^2 \cdot 3^2) - \frac{1}{2} \log_2 (3^4)\vphantom{\Bigl(}
   \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = \log_2 2^2 + \log_2 3^2 - \frac{1}{2} \log_2 3^4
   \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = 2 \log_2 2 + 2 \log_2 3 - \frac{1}{2} \cdot 4 \log_2 3
   \displaystyle \phantom{\log_2 36 - \frac{1}{2} \log_2 81}{} = 2\cdot 1 + 2 \log_2 3 - 2 \log_2 3 = 2\vphantom{\Bigl(}
4. \displaystyle \lg a^3 - 2 \lg a + \lg\frac{1}{a} = 3 \lg a - 2 \lg a + \lg a^{-1}
   \displaystyle \phantom{\lg a^3 - 2 \lg a + \lg\frac{1}{a}}{} = (3-2)\lg a + (-1) \lg a = \lg a - \lg a = 0

Example 9

1. Express \displaystyle \lg 5 as a natural logarithm.
   
   By definition, the \displaystyle \lg 5 is a number that satisfies the equality

   \displaystyle 10^{\lg 5} = 5\,\mbox{.}

   Take the natural logarithm ( ln ) of both sides.

   \displaystyle \ln 10^{\lg 5} = \ln 5\,\mbox{.}

   With the help of the logarithm law \displaystyle \ln a^b = b \ln a the left-hand side can be written as \displaystyle \lg 5 \cdot \ln 10 and the equality becomes

   \displaystyle \lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}

   Now divide both sides by \displaystyle \ln 10 giving the answer

   \displaystyle \lg 5 = \frac{\ln 5}{\ln 10} \qquad (\approx 0\textrm{.}699\,, \quad\text{dvs.}\ 10^{0\textrm{.}699} \approx 5)\,\mbox{.}

2. Express the 2-logarithm of 100 as a 10-logarithm lg.
   
   Using the definition of a logarithm one has that \displaystyle \log_2 100 formally satisfies

   \displaystyle 2^{\log_{\scriptstyle 2} 100} = 100

   and taking the 10-logarithm (lg) of both sides, one gets

   \displaystyle \lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}

   Since \displaystyle \lg a^b = b \lg a one gets \displaystyle \lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2 and the right-hand side can be simplified to \displaystyle \lg 100 = 2. This gives the equality

   \displaystyle \log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}

   Finally, dividing by \displaystyle \lg 2 gives that

   \displaystyle \log_{\scriptstyle 2} 100 = \frac{2}{\lg 2} \qquad ({}\approx 6\textrm{.}64\,, \quad\text{that is}\ 2^{6\textrm{.}64}\approx 100 )\,\mbox{.}

Example 10

1. Write \displaystyle 10^x using the base e.
   
   First, we write 10 as a power of e,

   \displaystyle 10 = e^{\ln 10}

   and then use the laws of exponents

   \displaystyle 10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10} \approx e^{2\textrm{.}3 x}\,\mbox{.}

2. Write \displaystyle e^{\,a} using the base 10.
   
   The number \displaystyle e can be written as \displaystyle e=10^{\lg e} and therefore

   \displaystyle e^a = (10^{\lg e})^a = 10^{\,a \cdot \lg e} \approx 10^{\,0\textrm{.}434a}\,\mbox{.}