expand each logarithm.
1) $\log \frac{2}{3}$
2) $\log (3 \cdot 11)$
3) $\log (6 \cdot 7)$
4) $\log (5 \cdot 11)$
5) $\log (7 \cdot 8)$
6) $\log \frac{12}{11}$
7) $\log \sqrt3{x}$
8) $\log \sqrt{x}$
9) $\log (a \cdot b)$
10) $\log x^5$
11) $\log (u \cdot v)^6$
12) $\log (ab^5)$

Question

Accepted Answer

### Problem 1: $\boldsymbol{\log \frac{2}{3}}$
# Explanation:
## Step1: Recall the quotient rule of logarithms: $\log \frac{a}{b} = \log a - \log b$ (where $a = 2$, $b = 3$)
$\log \frac{2}{3} = \log 2 - \log 3$
# Answer:
$\log 2 - \log 3$

### Problem 2: $\boldsymbol{\log (3 \cdot 11)}$
# Explanation:
## Step1: Recall the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = 3$, $b = 11$)
$\log (3 \cdot 11) = \log 3 + \log 11$
# Answer:
$\log 3 + \log 11$

### Problem 3: $\boldsymbol{\log (6 \cdot 7)}$
# Explanation:
## Step1: Apply the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = 6$, $b = 7$)
$\log (6 \cdot 7) = \log 6 + \log 7$
# Answer:
$\log 6 + \log 7$

### Problem 4: $\boldsymbol{\log (5 \cdot 11)}$
# Explanation:
## Step1: Use the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = 5$, $b = 11$)
$\log (5 \cdot 11) = \log 5 + \log 11$
# Answer:
$\log 5 + \log 11$

### Problem 5: $\boldsymbol{\log (7 \cdot 8)}$
# Explanation:
## Step1: Apply the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = 7$, $b = 8$)
$\log (7 \cdot 8) = \log 7 + \log 8$
# Answer:
$\log 7 + \log 8$

### Problem 6: $\boldsymbol{\log \frac{12}{11}}$
# Explanation:
## Step1: Use the quotient rule of logarithms: $\log \frac{a}{b} = \log a - \log b$ (where $a = 12$, $b = 11$)
$\log \frac{12}{11} = \log 12 - \log 11$
# Answer:
$\log 12 - \log 11$

### Problem 7: $\boldsymbol{\log \sqrt[3]{x}}$
# Explanation:
## Step1: Rewrite the cube root as an exponent: $\sqrt[3]{x} = x^{\frac{1}{3}}$
$\log \sqrt[3]{x} = \log x^{\frac{1}{3}}$
## Step2: Apply the power rule of logarithms: $\log a^n = n \log a$ (where $a = x$, $n = \frac{1}{3}$)
$\log x^{\frac{1}{3}} = \frac{1}{3} \log x$
# Answer:
$\frac{1}{3} \log x$

### Problem 8: $\boldsymbol{\log \sqrt{x}}$
# Explanation:
## Step1: Rewrite the square root as an exponent: $\sqrt{x} = x^{\frac{1}{2}}$
$\log \sqrt{x} = \log x^{\frac{1}{2}}$
## Step2: Use the power rule of logarithms: $\log a^n = n \log a$ (where $a = x$, $n = \frac{1}{2}$)
$\log x^{\frac{1}{2}} = \frac{1}{2} \log x$
# Answer:
$\frac{1}{2} \log x$

### Problem 9: $\boldsymbol{\log (a \cdot b)}$
# Explanation:
## Step1: Apply the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = a$, $b = b$)
$\log (a \cdot b) = \log a + \log b$
# Answer:
$\log a + \log b$

### Problem 10: $\boldsymbol{\log x^5}$
# Explanation:
## Step1: Use the power rule of logarithms: $\log a^n = n \log a$ (where $a = x$, $n = 5$)
$\log x^5 = 5 \log x$
# Answer:
$5 \log x$

### Problem 11: $\boldsymbol{\log (u \cdot v)^6}$
# Explanation:
## Step1: Apply the power rule of logarithms: $\log a^n = n \log a$ (where $a = u \cdot v$, $n = 6$)
$\log (u \cdot v)^6 = 6 \log (u \cdot v)$
## Step2: Use the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = u$, $b = v$)
$6 \log (u \cdot v) = 6 (\log u + \log v) = 6 \log u + 6 \log v$
# Answer:
$6 \log u + 6 \log v$

### Problem 12: $\boldsymbol{\log (ab^5)}$
# Explanation:
## Step1: Apply the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = a$, $b = b^5$)
$\log (ab^5) = \log a + \log b^5$
## Step2: Use the power rule of logarithms: $\log a^n = n \log a$ (where $a = b$, $n = 5$)
$\log b^5 = 5 \log b$
## Step3: Combine the results
$\log a + 5 \log b$
# Answer:
$\log a + 5 \log b$

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QUESTION IMAGE

Question

Explanation:

Problem 1: $\boldsymbol{\log \frac{2}{3}}$

Step1: Recall the quotient rule of logarithms: $\log \frac{a}{b} = \log a - \log b$ (where $a = 2$, $b = 3$)

Step1: Recall the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = 3$, $b = 11$)

Step1: Apply the product rule of logarithms: $\log (ab) = \log a + \log b$ (where $a = 6$, $b = 7$)

Answer:

Problem 2: $\boldsymbol{\log (3 \cdot 11)}$