3.3 Logarithms
From Förberedande kurs i matematik 1
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- | {{ | + | {{Selected tab|[[3.3 Logarithms|Theory]]}} |
- | {{ | + | {{Not selected tab|[[3.3 Exercises|Exercises]]}} |
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{{Info| | {{Info| | ||
- | ''' | + | '''Contents:''' |
* Logarithms | * Logarithms | ||
* Fundamental Laws of Logarithms | * Fundamental Laws of Logarithms | ||
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'''Learning outcomes:''' | '''Learning outcomes:''' | ||
- | After this section, you will have learned : | + | After this section, you will have learned: |
*The concepts of base and exponent. | *The concepts of base and exponent. | ||
*The meaning of the notation <math>\ln</math>, <math>\lg</math>, <math>\log</math> and <math>\log_{a}</math>. | *The meaning of the notation <math>\ln</math>, <math>\lg</math>, <math>\log</math> and <math>\log_{a}</math>. | ||
*To calculate simple logarithmic expressions using the definition of a logarithm. | *To calculate simple logarithmic expressions using the definition of a logarithm. | ||
*That logarithms are only defined for positive numbers. | *That logarithms are only defined for positive numbers. | ||
- | * | + | * The value of the number <math>e</math>. |
* To use the laws of logarithms to simplify logarithmic expressions. | * To use the laws of logarithms to simplify logarithmic expressions. | ||
* To know when the laws of logarithms are valid. | * To know when the laws of logarithms are valid. | ||
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==Logarithms to the base 10 == | ==Logarithms to the base 10 == | ||
- | We often use powers with base <math>10</math> to represent large and small numbers, for example | + | We often use powers with base <math>10</math> to represent large and small numbers, for example |
- | {{ | + | {{Displayed math||<math>\begin{align*} |
10^3 &= 10 \cdot 10 \cdot 10 = 1000\,,\\ | 10^3 &= 10 \cdot 10 \cdot 10 = 1000\,,\\ | ||
- | 10^{-2} &= \frac{1}{10 \cdot 10} = \frac{1}{100} = 0{ | + | 10^{-2} &= \frac{1}{10 \cdot 10} = \frac{1}{100} = 0\textrm{.}01\,\mbox{.} |
\end{align*}</math>}} | \end{align*}</math>}} | ||
- | If | + | If we only consider the exponents, we can informally state that |
- | :::"exponent | + | :::"the exponent of 1000 is 3", |
- | + | ||
- | This how logarithms are defined. | + | or |
+ | |||
+ | :::"the exponent of 0.01 is -2". | ||
+ | |||
+ | This is how logarithms are defined. We formalise this as follows: | ||
- | :::"''The logarithm'' of 1000 is 3", which is written as <math>\lg 1000 = 3</math>, | + | :::"''The logarithm'' of 1000 is 3", |
- | :::"''The logarithm'' of 0 | + | |
+ | which is written as <math>\lg 1000 = 3</math>, and | ||
+ | |||
+ | :::"''The logarithm'' of 0.01 is -2", | ||
+ | |||
+ | which is written as <math>\lg 0\textrm{.}01 = -2</math>. | ||
- | More generally, | + | More generally, we say: |
- | :::The logarithm of a number <math>y</math> is | + | :::The logarithm of a number <math>y</math> is denoted by <math>\lg y</math> and is the real number <math> x </math> in the blue box which satisfies the equality |
- | {{ | + | {{Displayed math||<math>10^{\ \bbox[#AAEEFF,2pt]{\,x\,}} = y\,\mbox{.} </math>}} |
- | Note that <math>y</math> must be a positive number for the logarithm <math>\lg y</math> to be defined, since there is no power of 10 that evaluates to a negative number or for that matter zero . | + | Note that <math>y</math> must be a positive number for the logarithm <math>\lg y</math> to be defined, since there is no power of 10 that evaluates to a negative number, or for that matter zero . |
<div class="exempel"> | <div class="exempel"> | ||
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10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}} | 10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}} | ||
= 100\,000</math>.</li> | = 100\,000</math>.</li> | ||
- | <li><math>\lg 0{ | + | <li><math>\lg 0\textrm{.}0001 = -4\quad</math> because <math> |
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} | ||
- | = 0{ | + | = 0\textrm{.}0001</math>.</li> |
<li><math>\lg \sqrt{10} = \frac{1}{2}\quad</math> because <math> | <li><math>\lg \sqrt{10} = \frac{1}{2}\quad</math> because <math> | ||
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}</math>.</li> | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}</math>.</li> | ||
- | <li><math>\lg 1 = 0\quad</math> | + | <li><math>\lg 1 = 0\quad</math> because <math> |
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li> | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li> | ||
<li><math>\lg 10^{78} = 78\quad</math> because <math> | <li><math>\lg 10^{78} = 78\quad</math> because <math> | ||
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}} | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}} | ||
= 10^{78}</math>.</li> | = 10^{78}</math>.</li> | ||
- | <li><math>\lg 50 \approx 1{ | + | <li><math>\lg 50 \approx 1\textrm{.}699\quad</math> because <math> |
- | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1{ | + | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\textrm{.}699\,}} \approx 50</math>.</li> |
<li><math>\lg (-10)</math> does not exist because <math> | <li><math>\lg (-10)</math> does not exist because <math> | ||
10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,a\vphantom{b,}\,}}</math> can never be -10 regardless of how <math>a</math> is chosen.</li> | 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,a\vphantom{b,}\,}}</math> can never be -10 regardless of how <math>a</math> is chosen.</li> | ||
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</div> | </div> | ||
- | In the penultimate example, one can easily understand that <math>\lg 50</math> must lie somewhere between 1 and 2 since <math>10^1 < 50 < 10^2</math>, | + | In the penultimate example, one can easily understand that <math>\lg 50</math> must lie somewhere between 1 and 2 since <math>10^1 < 50 < 10^2</math>. However, in practice to obtain a more precise value of the irrational number <math>\lg 50 = 1\textrm{.}69897\ldots</math> one needs a calculator (or table). |
<div class="exempel"> | <div class="exempel"> | ||
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== Different bases == | == Different bases == | ||
- | + | We can define logarithms with respect to a base other than 10. We must clearly indicate which number is used as the base for the logarithm. If we use a base such as 2 we write <math>\log_{\,2}</math> to mean a logarithm with base 2. | |
<div class="exempel"> | <div class="exempel"> | ||
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</div> | </div> | ||
- | + | We deal with logarithms which have other bases in the same way. | |
- | + | ||
<div class="exempel"> | <div class="exempel"> | ||
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= \frac{1}{16}</math>.</li> | = \frac{1}{16}</math>.</li> | ||
<li><math> \log_{\,b} \frac{1}{\sqrt{b}} = -\frac{1}{2}\quad | <li><math> \log_{\,b} \frac{1}{\sqrt{b}} = -\frac{1}{2}\quad | ||
- | </math> as | + | </math> as <math>b^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-1/2\,}} |
= \frac{1}{b^{1/2}} = \frac{1}{\sqrt{b}}</math> (if <math>b>0</math> and <math>b\not=1</math>).</li> | = \frac{1}{b^{1/2}} = \frac{1}{\sqrt{b}}</math> (if <math>b>0</math> and <math>b\not=1</math>).</li> | ||
</ol> | </ol> | ||
</div> | </div> | ||
- | If the base 10 is | + | If the base 10 is used, one rarely writes <math>\log_{\,10}</math>, but as we have previously seen we use the notation <math> \lg </math>, or simply <math>\log</math>. These symbols appear on many calculators. We should also note that it makes no sense to define logarithms with base <math>1</math>. |
==The natural logarithms == | ==The natural logarithms == | ||
- | In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2{ | + | In practice there are two bases that are commonly used for logarithms, 10 and the number <math>e</math> <math>({}\approx 2\textrm{.}71828 \ldots\,)</math>. Logarithms using the base ''e'' are called '' natural logarithms'' and we use the notation <math>\ln</math> instead of <math>\log_{\,e}</math>. |
<div class="exempel"> | <div class="exempel"> | ||
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<ol type="a"> | <ol type="a"> | ||
- | <li><math> \ln 10 \approx 2{ | + | <li><math> \ln 10 \approx 2{\textrm{.}}3\quad</math> because <math> |
- | e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{ | + | e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{\textrm{.}}3\,}} \approx 10</math>.</li> |
<li><math> \ln e = 1\quad</math> because <math> | <li><math> \ln e = 1\quad</math> because <math> | ||
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e</math>.</li> | e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e</math>.</li> | ||
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<li><math> \ln 1 = 0\quad</math> because <math> | <li><math> \ln 1 = 0\quad</math> because <math> | ||
e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li> | e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1</math>.</li> | ||
- | <li> | + | <li>If <math>y= e^{\,a}</math> then <math>a = \ln y</math>.</li> |
<li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5</math></li> | <li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5</math></li> | ||
<li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x</math></li> | <li><math> e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x</math></li> | ||
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</div> | </div> | ||
- | Most advanced calculators | + | Most advanced calculators have buttons for 10-logarithms and natural logarithms. |
== Laws of Logarithms == | == Laws of Logarithms == | ||
- | Between the years 1617 and 1624 Henry Biggs published a table of logarithms for all integers up to 20 000, and in 1628 Adriaan Vlacq expanded the table for all integers up to 100 000. The reason such an enormous amount of work was invested in producing these tables is that with the help of logarithms one can multiply numbers together just by adding their logarithms (addition | + | Between the years 1617 and 1624 Henry Biggs published a table of logarithms for all integers up to 20 000, and in 1628 Adriaan Vlacq expanded the table for all integers up to 100 000. The reason such an enormous amount of work was invested in producing these tables is that with the help of logarithms one can multiply numbers together just by adding their logarithms (addition is a much faster calculation than multiplication). |
<div class="exempel"> | <div class="exempel"> | ||
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<br> | <br> | ||
<br> | <br> | ||
- | If we know that <math>35 \approx 10^{\,1{ | + | If we know that <math>35 \approx 10^{\,1\textrm{.}5441}</math> and <math>54 \approx 10^{\,1\textrm{.}7324}</math> (i.e. <math>\lg 35 \approx 1\textrm{.}5441</math> and <math>\lg 54 \approx 1\textrm{.}7324</math>) then we can calculate that |
- | {{ | + | {{Displayed math||<math> |
- | 35 \cdot 54 \approx 10^{\,1{ | + | 35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324} |
- | = 10^{\,1{ | + | = 10^{\,1\textrm{.}5441 + 1\textrm{.}7324} |
- | = 10^{\,3{ | + | = 10^{\,3\textrm{.}2765}</math>.}} |
- | + | Since <math>10^{\,3\textrm{.}2765} \approx 1890</math> (i.e. <math>\lg 1890 \approx 3\textrm{.}2765</math>), we have thus managed to calculate the product | |
- | {{ | + | {{Displayed math||<math>35 \cdot 54 = 1890</math>}} |
- | + | just by adding together the exponents <math>1\textrm{.}5441</math> and <math>1\textrm{.}7324</math>. | |
</div> | </div> | ||
- | + | In the above example we have used a logarithmic law which states that | |
- | {{ | + | {{Displayed math||<math>\log (ab) = \log a + \log b</math>,}} |
- | + | for all <math>a,b>0</math>. We can see that this is true by using the laws of exponents. Indeed, | |
- | {{ | + | {{Displayed math||<math> |
a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b} | a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b} | ||
- | = \left\{ \mbox{laws of powers} \right\} | ||
= 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}</math>}} | = 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}}</math>}} | ||
and on the other hand, | and on the other hand, | ||
- | {{ | + | {{Displayed math||<math> |
a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}} | a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.}</math>}} | ||
- | By exploiting the laws of | + | so that we can see <math>\log (ab) = \log a + \log b</math>. By exploiting the laws of exponents in this way we can obtain the corresponding ''laws of logarithms'': |
<div class="regel"> | <div class="regel"> | ||
- | {{ | + | {{Displayed math||<math>\begin{align*} |
\log(ab) &= \log a + \log b,\\[4pt] | \log(ab) &= \log a + \log b,\\[4pt] | ||
\log\frac{a}{b} &= \log a - \log b,\\[4pt] | \log\frac{a}{b} &= \log a - \log b,\\[4pt] | ||
- | \log a^b &= b\cdot \log a\,\mbox{ | + | \log a^b &= b\cdot \log a\,\mbox{,}\\ |
\end{align*}</math>}} | \end{align*}</math>}} | ||
</div> | </div> | ||
- | + | for <math>a,b>0</math>. These laws hold regardless of the base we are working with. | |
- | + | ||
<div class="exempel"> | <div class="exempel"> | ||
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<div class="exempel"> | <div class="exempel"> | ||
- | ''' | + | '''Example 8''' |
<ol type="a"> | <ol type="a"> | ||
- | <li><math>\lg 9 + \lg 1000 - \lg 3 + \lg 0{ | + | <li><math>\lg 9 + \lg 1000 - \lg 3 + \lg 0\textrm{.}001 = \lg 9 + 3 - \lg 3 - 3 |
= \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3</math></li> | = \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3</math></li> | ||
<li><math>\ln\frac{1}{e} + \ln \sqrt{e} | <li><math>\ln\frac{1}{e} + \ln \sqrt{e} | ||
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== Changing the base == | == Changing the base == | ||
- | It sometimes | + | It is sometimes a good idea to express a logarithm as a logarithm with respect to another base. |
<div class="exempel"> | <div class="exempel"> | ||
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<br> | <br> | ||
<br> | <br> | ||
- | By definition, | + | By definition, <math>\lg 5</math> is a number that satisfies the equality |
- | {{ | + | {{Displayed math||<math>10^{\lg 5} = 5\,\mbox{.}</math>}} |
- | + | Taking the natural logarithm ( <math>\ln</math>) of both sides yields | |
- | {{ | + | {{Displayed math||<math>\ln 10^{\lg 5} = \ln 5\,\mbox{.}</math>}} |
- | With the help of the logarithm law <math>\ln a^b = b \ln a</math> the left-hand side can | + | With the help of the logarithm law <math>\ln a^b = b \ln a</math>, the left-hand side can be written as <math>\lg 5 \cdot \ln 10</math> and the equality becomes |
- | {{ | + | {{Displayed math||<math>\lg 5 \cdot \ln 10 = \ln 5\,\mbox{.}</math>}} |
- | Now divide both sides by <math>\ln 10</math> | + | Now divide both sides by <math>\ln 10</math> to get the answer |
- | {{ | + | {{Displayed math||<math> |
\lg 5 = \frac{\ln 5}{\ln 10} | \lg 5 = \frac{\ln 5}{\ln 10} | ||
- | \qquad (\approx 0{ | + | \qquad (\approx 0\textrm{.}699\,, |
- | \quad\text{dvs.}\ 10^{0{ | + | \quad\text{dvs.}\ 10^{0\textrm{.}699} \approx 5)\,\mbox{.} |
</math>}}</li> | </math>}}</li> | ||
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<br> | <br> | ||
Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies | Using the definition of a logarithm one has that <math>\log_2 100</math> formally satisfies | ||
- | {{ | + | {{Displayed math||<math>2^{\log_{\scriptstyle 2} 100} = 100</math>.}} |
- | + | Taking the 10-logarithm of both sides, we get | |
- | {{ | + | {{Displayed math||<math> |
\lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}} | \lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}</math>}} | ||
- | Since <math>\lg a^b = b \lg a</math> | + | Since <math>\lg a^b = b \lg a</math> we get <math>\lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2</math>, and moreover the right-hand side can be simplified to <math>\lg 100 = 2</math>. Thus we see that |
- | {{ | + | {{Displayed math||<math> |
\log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}} | \log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}</math>}} | ||
- | Finally dividing by <math>\lg 2</math> gives | + | Finally, dividing by <math>\lg 2</math> gives |
- | {{ | + | {{Displayed math||<math> |
\log_{\scriptstyle 2} 100 = \frac{2}{\lg 2} | \log_{\scriptstyle 2} 100 = \frac{2}{\lg 2} | ||
- | \qquad ({}\approx 6{ | + | \qquad ({}\approx 6\textrm{.}64\,, |
- | \quad\text{that is}\ 2^{6{ | + | \quad\text{that is}\ 2^{6\textrm{.}64}\approx 100 )\,\mbox{.}</math>}} |
</ol> | </ol> | ||
</div> | </div> | ||
- | The general formula for changing from one base <math>a</math> to another base <math>b</math> | + | The general formula for changing from one base <math>a</math> to another base <math>b</math> is |
- | {{ | + | {{Displayed math||<math> |
\log_{\scriptstyle\,a} x | \log_{\scriptstyle\,a} x | ||
= \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a} | = \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a} | ||
- | \,\mbox{ | + | \,\mbox{,}</math>}} |
- | + | and can be derived in the same way. We can also change the base of a power using logarithms. For instance, if we want to write <math> 2^5 </math> using the base 10 first write 2 as a power with the base 10: | |
- | {{ | + | {{Displayed math||<math>2 = 10^{\lg 2}</math>.}} |
- | + | Then, using one of the laws of exponents, | |
- | {{ | + | {{Displayed math||<math> |
2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2} | 2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2} | ||
- | \quad ({}\approx 10^{1 | + | \quad ({}\approx 10^{1\textrm{.}505}\,)\,\mbox{.}</math>}} |
<div class="exempel"> | <div class="exempel"> | ||
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<br> | <br> | ||
First, we write 10 as a power of ''e'', | First, we write 10 as a power of ''e'', | ||
- | {{ | + | {{Displayed math||<math>10 = e^{\ln 10}</math>.}} |
- | + | Using the the laws of exponents we can then see that | |
- | {{ | + | {{Displayed math||<math> |
10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10} | 10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10} | ||
- | \approx e^{2{ | + | \approx e^{2\textrm{.}3 x}\,\mbox{.}</math>}}</li> |
<li> Write <math>e^{\,a}</math> using the base 10. | <li> Write <math>e^{\,a}</math> using the base 10. | ||
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<br> | <br> | ||
The number <math>e</math> can be written as <math>e=10^{\lg e}</math> and therefore | The number <math>e</math> can be written as <math>e=10^{\lg e}</math> and therefore | ||
- | {{ | + | {{Displayed math||<math> |
e^a = (10^{\lg e})^a | e^a = (10^{\lg e})^a | ||
= 10^{\,a \cdot \lg e} | = 10^{\,a \cdot \lg e} | ||
- | \approx 10^{\,0{ | + | \approx 10^{\,0\textrm{.}434a}\,\mbox{.}</math>}} |
</ol> | </ol> | ||
</div> | </div> | ||
- | [[3.3 | + | [[3.3 Exercises|Exercises]] |
<div class="inforuta" style="width:580px;"> | <div class="inforuta" style="width:580px;"> | ||
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- | '''Keep in mind that | + | '''Keep in mind that... ''' |
- | You may need to spend | + | You may need to spend some time studying logarithms. |
- | Logarithms | + | Logarithms are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms. |
'''Reviews''' | '''Reviews''' | ||
- | For those of you who want to deepen your | + | For those of you who want to deepen your understanding or need more detailed explanations consider the following references: |
- | [http://en.wikipedia.org/wiki/Logarithm Learn more about logarithms | + | [http://en.wikipedia.org/wiki/Logarithm Learn more about logarithms on Wikipedia ] |
[http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/e.html Learn more about the number ''e'' in The MacTutor History of Mathematics archive ] | [http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/e.html Learn more about the number ''e'' in The MacTutor History of Mathematics archive ] | ||
- | ''' | + | '''Useful web sites''' |
[http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/LogGraph/logGraph.html Experiment with logarithms and powers ] | [http://www.ltcconline.net/greenl/java/IntermedCollegeAlgebra/LogGraph/logGraph.html Experiment with logarithms and powers ] |
Current revision
Theory | Exercises |
Contents:
- Logarithms
- Fundamental Laws of Logarithms
Learning outcomes:
After this section, you will have learned:
- The concepts of base and exponent.
- The meaning of the notation \displaystyle \ln, \displaystyle \lg, \displaystyle \log and \displaystyle \log_{a}.
- To calculate simple logarithmic expressions using the definition of a logarithm.
- That logarithms are only defined for positive numbers.
- The value of the number \displaystyle e.
- To use the laws of logarithms to simplify logarithmic expressions.
- To know when the laws of logarithms are valid.
- To express a logarithm in terms of a logarithm with a different base.
Logarithms to the base 10
We often use powers with base \displaystyle 10 to represent large and small numbers, for example
\displaystyle \begin{align*}
10^3 &= 10 \cdot 10 \cdot 10 = 1000\,,\\ 10^{-2} &= \frac{1}{10 \cdot 10} = \frac{1}{100} = 0\textrm{.}01\,\mbox{.} \end{align*} |
If we only consider the exponents, we can informally state that
- "the exponent of 1000 is 3",
or
- "the exponent of 0.01 is -2".
This is how logarithms are defined. We formalise this as follows:
- "The logarithm of 1000 is 3",
which is written as \displaystyle \lg 1000 = 3, and
- "The logarithm of 0.01 is -2",
which is written as \displaystyle \lg 0\textrm{.}01 = -2.
More generally, we say:
- The logarithm of a number \displaystyle y is denoted by \displaystyle \lg y and is the real number \displaystyle x in the blue box which satisfies the equality
\displaystyle 10^{\ \bbox[#AAEEFF,2pt]{\,x\,}} = y\,\mbox{.} |
Note that \displaystyle y must be a positive number for the logarithm \displaystyle \lg y to be defined, since there is no power of 10 that evaluates to a negative number, or for that matter zero .
Example 1
- \displaystyle \lg 100000 = 5\quad because \displaystyle 10^{\,\bbox[#AAEEFF,1pt]{\scriptstyle\,5\vphantom{,}\,}} = 100\,000.
- \displaystyle \lg 0\textrm{.}0001 = -4\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-4\vphantom{,}\,}} = 0\textrm{.}0001.
- \displaystyle \lg \sqrt{10} = \frac{1}{2}\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1/2\,}} = \sqrt{10}.
- \displaystyle \lg 1 = 0\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
- \displaystyle \lg 10^{78} = 78\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,78\vphantom{,}\,}} = 10^{78}.
- \displaystyle \lg 50 \approx 1\textrm{.}699\quad because \displaystyle 10^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\textrm{.}699\,}} \approx 50.
- \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.
In the penultimate example, one can easily understand that \displaystyle \lg 50 must lie somewhere between 1 and 2 since \displaystyle 10^1 < 50 < 10^2. However, in practice to obtain a more precise value of the irrational number \displaystyle \lg 50 = 1\textrm{.}69897\ldots one needs a calculator (or table).
Example 2
- \displaystyle 10^{\textstyle\,\lg 100} = 100
- \displaystyle 10^{\textstyle\,\lg a} = a
- \displaystyle 10^{\textstyle\,\lg 50} = 50
Different bases
We can define logarithms with respect to a base other than 10. We must clearly indicate which number is used as the base for the logarithm. If we use a base such as 2 we write \displaystyle \log_{\,2} to mean a logarithm with base 2.
Example 3
- \displaystyle \log_{\,2} 8 = 3\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 8.
- \displaystyle \log_{\,2} 2 = 1\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = 2.
- \displaystyle \log_{\,2} 1024 = 10\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,10\vphantom{,}\,}} = 1024.
- \displaystyle \log_{\,2}\frac{1}{4} = -2\quad because \displaystyle 2^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{2^2} = \frac{1}{4}.
We deal with logarithms which have other bases in the same way.
Example 4
- \displaystyle \log_{\,3} 9 = 2\quad because \displaystyle 3^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2\vphantom{,}\,}} = 9.
- \displaystyle \log_{\,5} 125 = 3\quad because \displaystyle 5^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,3\vphantom{,}\,}} = 125.
- \displaystyle \log_{\,4} \frac{1}{16} = -2\quad because \displaystyle 4^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-2\vphantom{,}\,}} = \frac{1}{4^2} = \frac{1}{16}.
- \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).
If the base 10 is used, one rarely writes \displaystyle \log_{\,10}, but as we have previously seen we use the notation \displaystyle \lg , or simply \displaystyle \log. These symbols appear on many calculators. We should also note that it makes no sense to define logarithms with base \displaystyle 1.
The natural logarithms
In practice there are two bases that are commonly used for logarithms, 10 and the number \displaystyle e \displaystyle ({}\approx 2\textrm{.}71828 \ldots\,). Logarithms using the base e are called natural logarithms and we use the notation \displaystyle \ln instead of \displaystyle \log_{\,e}.
Example 5
- \displaystyle \ln 10 \approx 2{\textrm{.}}3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,2{\textrm{.}}3\,}} \approx 10.
- \displaystyle \ln e = 1\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,1\vphantom{,}\,}} = e.
- \displaystyle \ln\frac{1}{e^3} = -3\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,-3\vphantom{,}\,}} = \frac{1}{e^3}.
- \displaystyle \ln 1 = 0\quad because \displaystyle e^{\scriptstyle\,\bbox[#AAEEFF,1pt]{\,0\vphantom{,}\,}} = 1.
- If \displaystyle y= e^{\,a} then \displaystyle a = \ln y.
- \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln 5\vphantom{,}\,}} = 5
- \displaystyle e^{\,\bbox[#AAEEFF,1pt]{\,\ln x\vphantom{,}\,}} = x
Most advanced calculators have buttons for 10-logarithms and natural logarithms.
Laws of Logarithms
Between the years 1617 and 1624 Henry Biggs published a table of logarithms for all integers up to 20 000, and in 1628 Adriaan Vlacq expanded the table for all integers up to 100 000. The reason such an enormous amount of work was invested in producing these tables is that with the help of logarithms one can multiply numbers together just by adding their logarithms (addition is a much faster calculation than multiplication).
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
\displaystyle
35 \cdot 54 \approx 10^{\,1\textrm{.}5441} \cdot 10^{\,1\textrm{.}7324} = 10^{\,1\textrm{.}5441 + 1\textrm{.}7324} = 10^{\,3\textrm{.}2765}. |
Since \displaystyle 10^{\,3\textrm{.}2765} \approx 1890 (i.e. \displaystyle \lg 1890 \approx 3\textrm{.}2765), we have thus managed to calculate the product
\displaystyle 35 \cdot 54 = 1890 |
just by adding together the exponents \displaystyle 1\textrm{.}5441 and \displaystyle 1\textrm{.}7324.
In the above example we have used a logarithmic law which states that
\displaystyle \log (ab) = \log a + \log b, |
for all \displaystyle a,b>0. We can see that this is true by using the laws of exponents. Indeed,
\displaystyle
a\cdot b = 10^{\textstyle\log a} \cdot 10^{\textstyle\log b} = 10^{\,\bbox[#AAEEFF,1pt]{\,\log a+\log b\,}} |
and on the other hand,
\displaystyle
a\cdot b = 10^{\,\bbox[#AAEEFF,1pt]{\,\log (ab)\,}}\,\mbox{.} |
so that we can see \displaystyle \log (ab) = \log a + \log b. By exploiting the laws of exponents in this way we can obtain the corresponding laws of logarithms:
\displaystyle \begin{align*}
\log(ab) &= \log a + \log b,\\[4pt] \log\frac{a}{b} &= \log a - \log b,\\[4pt] \log a^b &= b\cdot \log a\,\mbox{,}\\ \end{align*} |
for \displaystyle a,b>0. These laws hold regardless of the base we are working with.
Example 7
- \displaystyle \lg 4 + \lg 7 = \lg(4 \cdot 7) = \lg 28
- \displaystyle \lg 6 - \lg 3 = \lg\frac{6}{3} = \lg 2
- \displaystyle 2 \cdot \lg 5 = \lg 5^2 = \lg 25
- \displaystyle \lg 200 = \lg(2 \cdot 100) = \lg 2 + \lg 100 = \lg 2 + 2
Example 8
- \displaystyle \lg 9 + \lg 1000 - \lg 3 + \lg 0\textrm{.}001 = \lg 9 + 3 - \lg 3 - 3 = \lg 9- \lg 3 = \lg \displaystyle \frac{9}{3} = \lg 3
- \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(} - \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(} - \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
Changing the base
It is sometimes a good idea to express a logarithm as a logarithm with respect to another base.
Example 9
- Express \displaystyle \lg 5 as a natural logarithm.
By definition, \displaystyle \lg 5 is a number that satisfies the equality\displaystyle 10^{\lg 5} = 5\,\mbox{.} Taking the natural logarithm ( \displaystyle \ln) of both sides yields
\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 to get 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{.}
- 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. Taking the 10-logarithm of both sides, we get
\displaystyle \lg 2^{\log_{\scriptstyle 2} 100} = \lg 100\,\mbox{.}
Since \displaystyle \lg a^b = b \lg a we get \displaystyle \lg 2^{\log_2 100} = \log_{\scriptstyle 2} 100 \cdot \lg 2, and moreover the right-hand side can be simplified to \displaystyle \lg 100 = 2. Thus we see that
\displaystyle \log_{\scriptstyle 2} 100 \cdot \lg 2 = 2\,\mbox{.}
Finally, dividing by \displaystyle \lg 2 gives
\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{.}
The general formula for changing from one base \displaystyle a to another base \displaystyle b is
\displaystyle
\log_{\scriptstyle\,a} x = \frac{\log_{\scriptstyle\, b} x}{\log_{\scriptstyle\, b} a} \,\mbox{,} |
and can be derived in the same way. We can also change the base of a power using logarithms. For instance, if we want to write \displaystyle 2^5 using the base 10 first write 2 as a power with the base 10:
\displaystyle 2 = 10^{\lg 2}. |
Then, using one of the laws of exponents,
\displaystyle
2^5 = (10^{\lg 2})^5 = 10^{5\cdot \lg 2} \quad ({}\approx 10^{1\textrm{.}505}\,)\,\mbox{.} |
Example 10
- Write \displaystyle 10^x using the base e.
First, we write 10 as a power of e,\displaystyle 10 = e^{\ln 10}. Using the the laws of exponents we can then see that
\displaystyle 10^x = (e^{\ln 10})^x = e^{\,x \cdot \ln 10} \approx e^{2\textrm{.}3 x}\,\mbox{.}
- 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{.}
Study advice
The basic and final tests
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
Keep in mind that...
You may need to spend some time studying logarithms.
Logarithms are not usually dealt with in detail in high school. Therefore, many college students tend to encounter problems when it comes to calculations with logarithms.
Reviews
For those of you who want to deepen your understanding or need more detailed explanations consider the following references:
Learn more about logarithms on Wikipedia
Learn more about the number e in The MacTutor History of Mathematics archive
Useful web sites
Experiment with logarithms and powers
Help the frog to jump onto his water-lily leaf in the "log" game