Question

evaluate: \log_{27}9\
\bigcirc\\ \\ \\ \\ \frac{1}{3}\
\bigcirc\\ \\ \\ \\ \frac{2}{3}\
\bigcirc\\ \\ \\ \\ \frac{3}{2}\
\bigcirc\\ \\ \\ \\ -\frac{1}{3}\
done

Explanation:

Step1: Recall the change of base formula or logarithmic identity

We know that \( \log_{a^m}b^n=\frac{n}{m}\log_ab \), and also \( \log_{a}a = 1 \), \( \log_{a}b=\frac{\ln b}{\ln a} \) (change of base formula). Let's rewrite the numbers in terms of powers of 3. We know that \( 27 = 3^3 \) and \( 9=3^2 \). So we can use the property of logarithms \( \log_{a^m}a^n=\frac{n}{m} \).

Step2: Apply the logarithmic property

Given \( \log_{27}9=\log_{3^3}3^2 \). Using the property \( \log_{a^m}a^n = \frac{n}{m} \) (since \( \log_{a^m}a^n=\frac{\ln(a^n)}{\ln(a^m)}=\frac{n\ln a}{m\ln a}=\frac{n}{m} \) for \( a>0,a
eq1 \)), here \( a = 3 \), \( m = 3 \), \( n=2 \). So \( \log_{3^3}3^2=\frac{2}{3} \).

Answer:

\(\frac{2}{3}\) (corresponding to the option with \(\frac{2}{3}\))