$$ x $$ if $$ \log _{x} 8-\log _{x^{2}} 16=1 $$. $$ \log _{b} a=x $$, show that $$ b=a^{1 / x} $$ and deduce that olve, (i) $$ \log _{z^{2}} 4+\log _{4} n^{2}=3 $$ (ii) $$ 2^{2 x-1}+3 / 2=2^{x+1} $$ that $$ \log _{6} x=\frac{\log _{3} x}{1+\log 2} $$ hence given that $$ \log _{3} 2 $$ using tables or calculator $$ \log _{6} 4 $$. correct to o that $$ \log _{3} x=3 / 3 \log _{4} x $$ hence without using tab ate $$ \log _{3} 6_{3} $$ correct to 3 decimal places if 1 that $$ \log _{8} x=p $$ and $$ \log _{18} x=q $$, show that $$ \log $$ for $$ x $$ in the equation $$ \log _{4}(6-x)=\log _{2} x $$ the equation $$ \log _{2} x-\log _{x} \frac{4}{8}=2 $$

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UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Part 1: Solve $$ \log_{x} 8 - \log_{x^{2}} 16 = 1 $$

1. **Rewrite the logarithms**:
   $$
   \log_{x} 8 = \frac{\log 8}{\log x}
   $$
   $$
   \log_{x^{2}} 16 = \frac{\log 16}{\log x^{2}} = \frac{\log 16}{2 \log x}
   $$

2. **Substituting into the equation**:
   $$
   \frac{\log 8}{\log x} - \frac{\log 16}{2 \log x} = 1
   $$

3. **Combine the fractions**:
   $$
   \frac{2 \log 8 - \log 16}{2 \log x} = 1
   $$

4. **Multiply both sides by $$ 2 \log x $$**:
   $$
   2 \log 8 - \log 16 = 2 \log x
   $$

5. **Using properties of logarithms**:
   $$
   \log 8 = \log 2^3 = 3 \log 2 \quad \text{and} \quad \log 16 = \log 2^4 = 4 \log 2
   $$
   Substitute these into the equation:
   $$
   2(3 \log 2) - 4 \log 2 = 2 \log x
   $$
   $$
   6 \log 2 - 4 \log 2 = 2 \log x
   $$
   $$
   2 \log 2 = 2 \log x
   $$

6. **Dividing both sides by 2**:
   $$
   \log 2 = \log x
   $$

7. **Exponentiating both sides**:
   $$
   x = 2
   $$

### Part 2: Show that $$ b = a^{1/x} $$ if $$ \log_{b} a = x $$

1. **Starting from the definition of logarithm**:
   $$
   \log_{b} a = x \implies b^x = a
   $$

2. **Rearranging gives**:
   $$
   b = a^{1/x}
   $$

### Part 3: Solve $$ \log_{z^{2}} 4 + \log_{4} n^{2} = 3 $$

1. **Rewrite the logarithms**:
   $$
   \log_{z^{2}} 4 = \frac{\log 4}{\log z^{2}} = \frac{\log 4}{2 \log z}
   $$
   $$
   \log_{4} n^{2} = \frac{\log n^{2}}{\log 4} = \frac{2 \log n}{\log 4}
   $$

2. **Substituting into the equation**:
   $$
   \frac{\log 4}{2 \log z} + \frac{2 \log n}{\log 4} = 3
   $$

3. **Multiply through by $$ 2 \log z \log 4 $$** to eliminate the denominators:
   $$
   \log 4 \cdot \log 4 + 4 \log n \cdot \log z = 6 \log z \cdot \log 4
   $$

4. **Rearranging gives**:
   $$
   4 \log n \cdot \log z = 6 \log z \cdot \log 4 - \log^2 4
   $$

### Part 4: Solve $$ 2^{2x-1} + \frac{3}{2} = 2^{x+1} $$

1. **Rearranging the equation**:
   $$
   2^{2x-1} - 2^{x+1} + \frac{3}{2} = 0
   $$

2. **Let $$ y = 2^x $$**, then $$ 2^{2x} = y^2 $$:
   $$
   \frac{y^2}{2} - 2y + \frac{3}{2} = 0
   $$

3. **Multiply through by 2**:
   $$
   y^2 - 4y + 3 = 0
   $$

4. **Factoring gives**:
   $$
   (y - 1)(y - 3) = 0
   $$

5. **Thus, $$ y = 1 $$ or $$ y = 3 $$**:
   $$
   2^x = 1 \implies x = 0
   $$
   $$
   2^x = 3 \implies x = \log_2 3
   $$

### Part 5: Show that $$ \log_{6} x = \frac{\log_{3} x}{1 + \log 2} $$

1. **Using the change of base formula**:
   $$
   \log_{6} x = \frac{\log_{3} x}{\log_{3} 6}
   $$

2. **Express $$ \log_{3} 6 $$**:
   $$
   \log_{3} 6 = \log_{3} (2 \cdot 3) = \log_{3} 2 + 1
   $$

3. **Thus**:
   $$
   \log_{6} x = \frac{\log_{3} x}{\log_{3} 2 + 1}
   $$

### Part 6: Calculate $$ \log_{6} 4 $$ using $$ \log_{3} 2 $$

1. **Using the previous result**:
   $$
   \log_{6} 4 = \frac{\log_{3} 4}{\log_{3} 6} = \frac{2 \log_{3} 2}{\log_{3} 2 + 1}
   $$

### Part 7: Solve $$ \log_{4}(6-x) = \log_{2} x $$

1. **Using properties of logarithms**:
   $$
   \log_{4}(6-x) = \frac{\log_{2}(6-x)}{2}
   $$
   $$
   \frac{\log_{2}(6-x)}{2} = \log_{2} x
   $$

2. **Multiply through by 2**:
   $$
   \log_{2}(6-x) = 2 \log_{2} x
   $$

3. **Exponentiating gives**:
   $$
   6 - x = x^2
   $$
   \
$$ x = 2 $$

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