How do you evaluate ${log}_{27} (\frac{1}{3})$ $log_27 (1/3)$ ?

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How do you evaluate ${log}_{27} (\frac{1}{3})$ $log_27 (1/3)$ ?

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Answer 1 · 2016-08-04T06:26:19

${log}_{27} (\frac{1}{3}) = - \frac{1}{3}$ $log_27 (1/3) = -1/3$

Explanation:

By definition $log$ $log$ is the inverse of exponentiation and vice versa for any base $b$ $b$ . That is:

${log}_{b} (b^{x}) = x$ $log_b (b^x) = x$ for all Real values of $x$ $x$

$b^{{log}_{b} x} = x$ $b^(log_b x) = x$ for all $x > 0$ $x > 0$

The change of base formula tells us that for any $a, b, c > 0$ $a, b, c > 0$ with $a \neq 1$ $a != 1$ and $c \neq 1$ $c != 1$ :

${log}_{a} b = \frac{{log}_{c} b}{{log}_{c} a}$ $log_a b = (log_c b) / (log_c a)$

So we find:

${log}_{27} (\frac{1}{3}) = \frac{{log}_{3} (\frac{1}{3})}{{log}_{3} 27} = \frac{{log}_{3} 3^{- 1}}{{log}_{3} 3^{3}} = - \frac{1}{3}$ $log_27 (1/3) = (log_3 (1/3)) / (log_3 27) = (log_3 3^(-1)) / (log_3 3^3) = -1/3$

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How do you evaluate ${log}_{27} (\frac{1}{3})$ $log_27 (1/3)$ ?

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