Simplify the expression $4\log_3 2 - \log_3 4 - 2\log_3 \sqrt{3} - \log_3 12$ and write the final answer as a single logarithm.

1. Problem Description

Simplify the expression $4\log_3 2 - \log_3 4 - 2\log_3 \sqrt{3} - \log_3 12$ and write the final answer as a single logarithm.

2. Solution Steps

First, we use the power rule of logarithms: $\log_b(x^n) = n \log_b x$ to rewrite the expression.

$4\log_3 2 = \log_3 (2^4) = \log_3 16$

$2\log_3 \sqrt{3} = \log_3 (\sqrt{3})^2 = \log_3 3 = 1$

The expression becomes:

$\log_3 16 - \log_3 4 - \log_3 3 - \log_3 12$

Next, we use the quotient rule of logarithms: $\log_b x - \log_b y = \log_b \frac{x}{y}$.

$\log_3 16 - \log_3 4 = \log_3 \frac{16}{4} = \log_3 4$

$\log_3 4 - \log_3 3 - \log_3 12 = \log_3 \frac{4}{3} - \log_3 12 = \log_3 \frac{4/3}{12} = \log_3 \frac{4}{3 \cdot 12} = \log_3 \frac{4}{36} = \log_3 \frac{1}{9}$

Since $9 = 3^2$, then $\frac{1}{9} = 3^{-2}$.

$\log_3 \frac{1}{9} = \log_3 3^{-2}$

Using the power rule again,

$\log_3 3^{-2} = -2 \log_3 3 = -2 \cdot 1 = -2$

We can rewrite $-2$ as a single logarithm:

$-2 = \log_3 3^{-2} = \log_3 \frac{1}{9}$

Alternatively,

$\log_3 16 - \log_3 4 - \log_3 3 - \log_3 12 = \log_3 \frac{16}{4 \cdot 3 \cdot 12} = \log_3 \frac{16}{144} = \log_3 \frac{1}{9} = \log_3 3^{-2} = -2 \log_3 3 = -2$

We want to write this as a single logarithm with base

3. We can do so by expressing $-2$ as $\log_3 (3^{-2})$.

$\log_3 (3^{-2}) = \log_3 \frac{1}{9}$

3. Final Answer

$\log_3 \frac{1}{9}$