Question 235481
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The best I can do without a calculator is to express this in terms of a single logarithm.


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n) = n\log_b(x)]


To write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\log_4(6)\ -\ \log_4(5)\ =\ \log_4(6^4)\ -\ \log_4(5)]


Then use: <i>The difference of the logs is the log of the quotient</i>, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = log_b\left(\frac{x}{y}\right)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\log_4(6)\ -\ \log_4(5)\ =\ \log_4(6^4)\ -\ \log_4(5)\ =\ \log_4\left(\frac{6^4}{5}\right)]


I need at least a scratch pad to go any further: *[tex \Large 6\times6=36], *[tex \Large 36\times36=1296], *[tex \Large 1296\div5=259.2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\log_4(6)\ -\ \log_4(5)\ =\ \log_4(6^4)\ -\ \log_4(5)\ =\ \log_4\left(\frac{6^4}{5}\right)\ =\ \log_4(259.2)]


Now, if you want to do this the old fashioned way with a table of logarithms, you aren't likely to find a table of base 4 logs.  That means you will have to do a base conversion.


Here's the formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x)\ =\ \frac{\log_c(x)}{\log_c(b)}]


So, for your situation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4(259.2)\ =\ \frac{\log_{10}(259.2)}{\log_{10}(4)}]


Then you would have to look up *[tex \Large \log_{10}(2.59)] and *[tex \Large \log_{10}(2.60)], then interpolate to find *[tex \Large \log_{10}(2.592)].


To interpolate:  *[tex \Large \log_{10}(2.59)\ + .2\left(\log_{10}(2.60)\ - \log_{10}(2.59)\right)].


Finally, calculate *[tex \Large \log_{10}(2.592)\ +\ 2\ =\ \log_{10}(259.2)].


Next, look up *[tex \Large \log_{10}(4)] and divide it into the results of the last step -- by hand, of course.


Good luck.


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