To rewrite the logarithm [tex]\(\log_4\left(\frac{1}{16}mn^2\right)\)[/tex] as a sum or difference of logarithms, we must use logarithmic properties, such as the logarithm of a quotient and the logarithm of a product. Recall the following properties:
1. [tex]\(\log_b\left(\frac{A}{B}\right) = \log_b(A) - \log_b(B)\)[/tex]
2. [tex]\(\log_b(AB) = \log_b(A) + \log_b(B)\)[/tex]
3. [tex]\(\log_b(A^k) = k \log_b(A)\)[/tex]
Given [tex]\(\log_4\left(\frac{1}{16}mn^2\right)\)[/tex]:
### Step 1: Apply the Quotient Rule
[tex]\[
\log_4\left(\frac{1}{16}mn^2\right) = \log_4(1) - \log_4(16mn^2)
\][/tex]
### Step 2: Evaluate [tex]\(\log_4(1)\)[/tex]
[tex]\[
\log_4(1) = 0 \quad \text{(since any number's logarithm of 1 is 0)}
\][/tex]
So the expression simplifies to:
[tex]\[
0 - \log_4(16mn^2) = -\log_4(16mn^2)
\][/tex]
### Step 3: Use the Product Rule
[tex]\[
-\log_4(16mn^2) = -\left(\log_4(16) + \log_4(m) + \log_4(n^2)\right)
\][/tex]
### Step 4: Simplify [tex]\(\log_4(16)\)[/tex] and [tex]\(\log_4(n^2)\)[/tex]
Notice that [tex]\(16 = 4^2\)[/tex], so:
[tex]\[
\log_4(16) = \log_4(4^2) = 2 \log_4(4) = 2 \cdot 1 = 2 \quad \text{(since } \log_4(4) = 1\text{)}
\][/tex]
Now simplify [tex]\(\log_4(n^2)\)[/tex]:
[tex]\[
\log_4(n^2) = 2 \log_4(n)
\][/tex]
### Step 5: Substitute the simplified terms back in
[tex]\[
-\left(\log_4(16) + \log_4(m) + \log_4(n^2)\right) = -(2 + \log_4(m) + 2 \log_4(n))
\][/tex]
### Step 6: Distribute the negative sign
[tex]\[
-(2 + \log_4(m) + 2 \log_4(n)) = -2 - \log_4(m) - 2 \log_4(n)
\][/tex]
Thus, the simplified form of [tex]\(\log_4\left(\frac{1}{16}mn^2\right)\)[/tex] in terms of a sum or difference of logarithms is:
[tex]\[
-2 - \log_4(m) - 2 \log_4(n)
\][/tex]
