Question 238097
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There are two ways to do this one.


<b><i>Method 1: Definition of Logarithms</i></b>


Use the fact that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


To write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_8(\frac{1}{4}) \ \ \Rightarrow\ \ 8^y = \frac{1}{4}]


But since *[tex \Large 8\ =\ 2^3] we can say *[tex \Large 8^y\ =\ 2^{3y}] and since *[tex \Large 4\ =\ 2^2] we can say *[tex \Large \frac{1}{4}\ =\ 2^{-2}]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{3y}\ =\ 2^{-2}]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3y\ =\ -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{2}{3}]


<b><i>Method 2: Base Conversion</i></b>


Calculators and Log Tables do not support direct look up of logs other than base 10 or base e (natural) logs.  But you can  do a base conversion:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log _b \left( x \right)\ =\ \frac{{\log _c \left( x \right)}}{{\log _c \left( b \right)}}]


Where *[tex \Large b\ =\ 8], *[tex \Large c\ =\ 10] or *[tex \Large c\ =\ e] (it doesn't matter -- you can use either), and *[tex \Large x\ =\ \frac{1}{4}]


So, using *[tex \Large c\ =\ e]:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log _8 \left( \frac{1}{4} \right)\ =\ \frac{{\ln \left( \frac{1}{4} \right) }}{{\ln \left( 8 \right)}}]


Use your calculator or a natural log table to fill in the values of *[tex \Large \ln \left( \frac{1}{4} \right) ] and *[tex \Large \ln \left( 8 \right)], and then do the quotient to calculate the value.


Go here for an on-line natural log table:  http://myhandbook.info/table_naperlog.html


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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