what is the value of $log_{27}9$?\n- $\frac{3}{2}$\n- $\frac{2}{3}$\n- $\frac{2}{3}$\n- $\frac{3}{2}$

Explanation:

Step1: Recall the change - of - base formula

Let (y = \log_{27}9). By the change - of - base formula (\log_{a}b=\frac{\log_{c}b}{\log_{c}a}), we can write (y=\frac{\log 9}{\log 27}).

Step2: Express 9 and 27 as powers of 3

Since (9 = 3^{2}) and (27=3^{3}), then (\log 9=\log(3^{2}) = 2\log 3) and (\log 27=\log(3^{3})=3\log 3).

Step3: Substitute into the fraction

Substitute (\log 9 = 2\log 3) and (\log 27 = 3\log 3) into (y=\frac{\log 9}{\log 27}), we get (y=\frac{2\log 3}{3\log 3}).

Step4: Simplify the fraction

Cancel out (\log 3) in the numerator and denominator, so (y = \frac{2}{3}).

Answer:

C. (\frac{2}{3})