Question

write the expression below as a single logarithm in simplest form.\\(\log_{b} 9 - 2\log_{b} 3\\)

Explanation:

Step1: Apply power rule of logarithms

The power rule states that \( n\log_{b}a=\log_{b}a^{n} \). So, for \( 2\log_{b}3 \), we have \( 2\log_{b}3 = \log_{b}3^{2}=\log_{b}9 \).

Step2: Substitute and simplify

Substitute \( 2\log_{b}3=\log_{b}9 \) into the original expression \( \log_{b}9 - 2\log_{b}3 \), we get \( \log_{b}9-\log_{b}9 \).
Since \( \log_{b}x-\log_{b}y = \log_{b}\frac{x}{y} \), here \( x = 9,y = 9 \), so \( \log_{b}\frac{9}{9}=\log_{b}1 \). And we know that \( \log_{b}1 = 0 \) for any valid base \( b>0,b
eq1 \).

Answer:

\( 0 \)