##### The function f(x) = log9 x is the logarithm function with base . So f(9) =

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The function
f(x) = log9 x
is the logarithm function with base  . So f(9) =  , f(1) =  ,
f
 1 9

=  , f(81) =  , and f(3) =  .

Jul 16th, 2015

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$\fn_jvn log_ba=c\hspace{5}means\hspace{5}a=b^{c}$

For example,

$\fn_jvn log_28=3\hspace{5}means\hspace{5}8=2^{3}$

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The function

$\fn_jvn f(x)=log_9x$

is the logarithm function with base 9

$\fn_jvn f(9)=log_99=1\hspace{5}(because\hspace{5}9^1=9)$

$\fn_jvn f(1)=log_91=0\hspace{5}(because\hspace{5}9^0=1)$

$\fn_jvn \\ f(1/9)=log_9\frac{1}{9}\\ \\ f(1/9)=log_99^{-1}\\ \\ f(1/9)=-1log_99\hspace{5}(because\hspace{5}log_nx^m=mlog_nx)\\ \\ f(1/9)=-1\times1\hspace{5}(because\hspace{5}9^1=9)\\ \\ f(1/9)=-1$

$\fn_jvn \\ f(81)=log_981\\ \\ f(81)=log_99^2\\ \\ f(81)=2log_99\hspace{5}(because\hspace{5}log_nx^m=mlog_nx)\\ \\ f(81)=2\times1\hspace{5}(because\hspace{5}9^1=9)\\ \\ f(81)=2$

$\fn_jvn \\ f(3)=log_93\\ \\ f(3)=log_99^{1/2}\hspace{5}(because\hspace{5}\sqrt{9}=9^{1/2}=3)\\ \\ f(3)=\frac{1}{2}log_99\hspace{5}(because\hspace{5}log_nx^m=mlog_nx)\\ \\ f(3)=\frac{1}{2}\times1\hspace{5}(because\hspace{5}9^1=9)\\ \\ f(3)=\frac{1}{2}\\ \\ f(3)=0.5$

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Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 17th, 2015

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Jul 16th, 2015
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Jul 16th, 2015
Oct 19th, 2017
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