How do you evaluate ${log}_{27} 9$ $log_27 9$ ?

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Answer 1 · 2016-08-07T18:16:41

$x = \frac{2}{3}$ $x = 2/3$

In log form the question being asked is ...

"To what power must 27 be raised to give 9?#

You should recognise 27 as $3^{3}$ $3^3$ and 9 as $3^{2}$ $3^2$

$3 = \sqrt[3]{27} = 27^{\frac{1}{3}}$ $3 = root3(27) = 27^(1/3)$

$9 = 3^{2} = {(\sqrt[3]{27})}^{2} = 27^{{(\frac{1}{3})}^{2}} = 27^{\frac{2}{3}}$ $9 = 3^2 = (root3(27))^2 = 27^((1/3)^2) = 27^(2/3)$

OR

${log}_{27} 9 = x \Rightarrow 27^{x} = 9$ $log_27 9 = x rArr " " 27^x = 9$

$3^{3 x} = 3^{2}$ $3^(3x) = 3^2$

Equating exponents: $3 x = 2$ $3x = 2$

$x = \frac{2}{3}$ $x = 2/3$

How do you evaluate log279log_27 9?