04 1logarithms

Logarithms

In the equation $y = a^x$,

  • x is the index (or power or exponent),
  • The index, x, is also called a logarithm.

$y = a^x$ is sometimes called an indicial equation (or exponential equation) because the variable is in the index (or exponent).

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Writing an Exponential as a Log

$y=a^x$ can also be written as $\log_a \, (y) =x$ … … (provided a > 0)

For example:
… … $8 = 2^3$ can also be written as $\log_2 \, (8) = 3$

  • this means that with a base of 2, 3 is the index needed to give an answer of 8

For example:
… … $\log_{10} \, (73) = 1.863$ is telling us that $10^{1.863} = 73$

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NOTE: $\log_a(y)$ is only defined when y > 0

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The equivalence between the two equations is shown using the double headed arrow:

… … $10^4 = 10,000 \; \iff \; \log_{10}\,(10,000) = 4$

  • The double headed arrow is notation for iff, which means "if and only if"

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Logarithm Laws

Each of the Index Laws has a corresponding Log Law
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… …1$\log_a \,(mn) = \log_a \,(m) + \log_a \, (n)$
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… … 2$\log_a \, \big(\frac{m}{n}\big) = \log_a\, (m) - \log_a\, (n)$
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… … 3$\log_a \, \big( m^p \big) = p \, \log_a\, (m)$
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… … 4$\log_a \, (1) = 0$
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… … 5$\log_a \, \big(\frac{1}{m}\big) = - \log_a \, (m)$
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… … 6$\log_a\, (a) = 1$

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Change of Base Rule

It is often useful to change the base of a logarithmic expression.

… … $\log_a\,(y) = \dfrac{\log_b\,(y)}{\log_b\,(a)}$

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For example
… … $\log_2 \, (x) = \dfrac{\log_{10} \, (x)}{\log_{10}\,(2)}$

… … This changes the expression from base 2, to base 10

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Example 1

Write the following in index form:

… … a)$\log_{10}(1000)=3$

… … b)$\log_x (81)=4$

… … c)$\log_5 (x)=4$

Solution:

… … a)$\log_{10}(1000)=3 \; \; \iff \; \;10^3=1000$

… … b)$\log_x(81)=4 \;\; \iff \;\; x^4=81$

… … c)$\log_5 (x)=4 \;\; \iff \;\; 5^4=x$

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Example 2

Use $a^x=y \;\; \iff \;\; \log_a (y)=x$ to simplify the following:

… … a)$\log_4 (64)$

… … b)$\log_3\big(\frac{1}{27}\big)$
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Solution:

… … a) … Let $\log_4 (64)=x$

… … … .. then $\log_4 (64)=x \;\; \iff \;\; 4^x=64$

… … … … … $4^x=4^3$

… … … … … $x=3$

… … … … … so

… … … … … $\log_4 (64) = 3$
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NOTE: We could have also simplified this by using log laws:

… … a)$\log_4 (64)$

… … … .. $= \log_{4} \big( 4^3 \big)$

… … … .. $= 3\log_{4} (4)$ … … {using law 3 from above}

… … … .. $= 3$ … … … {using law 6 from above}
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… … b) … Let $\log_3\big(\frac{1}{27}\big)=x$

… … … .. then $\log_3\big(\frac{1}{27}\big) = x \;\; \iff \;\; 3^x = \frac{1}{27}$

… … … … … $3^x = \dfrac{1}{3^3}$

… … … … … $3^x = 3^{-3}$

… … … … … $x =-3$

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Example 3

… … Simplify $\log_4 (6) - \log_4 (96)$

Solution:

… … $\log_4(6) - \log_4 (96)$

… … $= \log_4\Bigg( \dfrac{6}{96} \Bigg)$ … … {using law 2 from above}

… … $= \log_4 \Bigg( \dfrac{1}{16} \Bigg)$

… … $=\log_4\Bigg(\dfrac{1}{4^2}\Bigg)$

… … $=\log_4 \Big( 4^{-2} \Big)$

… … $=-2\log_4 (4)$ … … {using law 3 from above}

… … $= -2$ … … … {using law 6 from above}

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Example 4

… … Simplify $\dfrac{\log_2\,(8)}{log_2\, (64)}$

Solution:

… … $\dfrac{\log_2 \,(8)}{\log_2 \, (64)}$

… … $= \dfrac{\log_2 \, \big(2^3 \big) }{\log_2 \, \big( 2^6 \big) }$

… … $= \dfrac{3\log_2 \,(2)}{6\log_2 (2) }$ … … {using law 3 from above}

… … $=\dfrac{3}{6}$ … … …{because $\log_2 (2)$ is a common factor}

… … $=\dfrac{1}{2}$

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Your calculator can simplify logs:

04.1Example4b.gif

Example 5

… … Express the following in the form Alogn(B)

… … $5\log_3 (x) + 3- \log_3 \big(x^2 \big)$

Solution:

… … $5\log_3 (x)+3-\log_3 \big( x^2 \big)$

… … $=\log_3 \big( x^5 \big) + 3\log_3 (3) - \log_3 \big(x^2 \big)$ … …{using law 6}

… … $= \log_3 \big(x^5 \big) + \log_3 \big(3^3 \big) - \log_3 \big( x^2 \big)$ … … {using law 3}

… … $= \log_3 \Bigg( \dfrac{x^5 \times 3^3}{x^2} \Bigg)$ … … {using law 1 and 2}

… … $= \log_3 \Big( (3x)^3 \Big)$

… … $= 3\log_3 (3x)$ … … … {using law 6}

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For another site that explains this idea, go here: MathsIsFun

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