Question

question
write the expression below as a single logarithm in simplest form.
\\(\log_b 9 - 2\log_b 3\\)
this is the only question in this section.
watch video
show examples

Explanation:

Step1: Use power rule of logarithms

The power rule of logarithms states that \( n\log_b a=\log_b a^n \). So, for \( 2\log_b 3 \), we can rewrite it as \( \log_b 3^2 \). Since \( 3^2 = 9 \), this becomes \( \log_b 9 \).
So the expression \( \log_b 9 - 2\log_b 3 \) becomes \( \log_b 9-\log_b 9 \).

Step2: Use quotient rule of logarithms

The quotient rule of logarithms states that \( \log_b a-\log_b c=\log_b \frac{a}{c} \). Here, \( a = 9 \) and \( c=9 \), so \( \log_b 9-\log_b 9=\log_b \frac{9}{9} \).

Step3: Simplify the argument

Simplify \( \frac{9}{9}=1 \), so we have \( \log_b 1 \). And we know that \( \log_b 1 = 0 \) for any valid base \( b>0,b
eq1 \), but also, we can check the steps again. Wait, actually, when we apply the quotient rule:

Wait, let's re - do step 1 and step 2 correctly.

Step1: Apply power rule

Given \( 2\log_b 3 \), by power rule \( n\log_b x=\log_b x^n \), so \( 2\log_b 3=\log_b 3^2=\log_b 9 \).

Step2: Apply subtraction rule of logarithms

The subtraction rule of logarithms is \( \log_b M-\log_b N=\log_b\frac{M}{N} \). So \( \log_b 9 - \log_b 9=\log_b\frac{9}{9}=\log_b 1 \). And since \( \log_b 1 = 0 \) (because \( b^0 = 1 \) for \( b>0,b
eq1 \)), but also, we can see that the original expression is \( \log_b 9-2\log_b 3=\log_b 9-\log_b 9 = 0 \), and we can also express it as a single logarithm \( \log_b 1 \) (since \( \log_b 1=0 \)).

Answer:

\(\log_{b}1\) (or \(0\), since \(\log_{b}1 = 0\) for \(b>0,b
eq1\))