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logbase3(1/9)=x find x please explain
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can you use equation it is hard to understand
but the answer can be -2
\[\log_{3}(1/9)=x \]
No answer is -2
x=logbase3(1/9) = logbase3(1) - logbase3(9) = 0 - logbase3(3^2) = - 2logbase3(3) = -2
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\[3^1=3\]\[3^0=1\]\[3^{-1}=\frac13\]\[3^{-2}=\frac19\]in general\[a^{-n}=\frac1{a^n}\]
then \[\log_{3} (3^{-2}) \] \[-2\log_{3} (3)\] -2
\[\log_3(\frac19)\]is like asking "what power do you raise 3 to to get 1/9?" the answer is -2
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