(1) Find the value of $$ \log 2512 $$ ? (2) Expand $$ \log \frac{x 2 y 3}{z^{2}} $$ (3) If $$ 5^{x+1}=7^{1-x} $$ Find $$ X $$ value. (4) If $$ x^{2}+y^{2}=27 x y $$, Show that $$ \log \left(\frac{x-y}{5}\right)=\frac{1}{2}(\log x+\log y) $$

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Let's solve each of the problems step-by-step.

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### **Problem (1): Find the value of $$ \log 2512 $$.**

Assuming the logarithm is base 10:

1. **Estimate Using Known Log Values:**
   - $$ \log(1000) = 3 $$
   - $$ \log(2000) \approx 3.3010 $$ (since $$ \log(2) \approx 0.3010 $$)
   - $$ \log(2500) \approx 3.3979 $$ (since $$ \log(25) \approx 1.3979 $$ and $$ \log(100) = 2 $$)

2. **Refine the Estimate for $$ \log(2512) $$:**
   - The difference between 2512 and 2500 is 12.
   - Use a linear approximation:
     $$
     \log(2512) \approx \log(2500) + \frac{12}{2500 \times \ln(10)} \approx 3.3979 + 0.0021 = 3.4000
     $$

**Answer:**  
$$ \log 2512 \approx 3.4000 $$

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### **Problem (2): Expand $$ \log \frac{x^2 y^3}{z^{2}} $$.**

Use logarithm properties to expand the expression:

$$
\log \left( \frac{x^2 y^3}{z^2} \right) = \log(x^2) + \log(y^3) - \log(z^2)
$$

Apply the power rule ($$ \log(a^n) = n \log a $$):

$$
= 2\log x + 3\log y - 2\log z
$$

**Answer:**  
$$ 2 \log x + 3 \log y - 2 \log z $$

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### **Problem (3): If $$ 5^{x+1}=7^{1-x} $$, find the value of $$ x $$.**

1. **Take the Natural Logarithm of Both Sides:**

$$
\ln(5^{x+1}) = \ln(7^{1-x})
$$
$$
(x + 1) \ln 5 = (1 - x) \ln 7
$$

2. **Expand and Rearrange Terms:**

$$
x \ln 5 + \ln 5 = \ln 7 - x \ln 7
$$
$$
x (\ln 5 + \ln 7) = \ln 7 - \ln 5
$$

3. **Solve for $$ x $$:**

$$
x = \frac{\ln 7 - \ln 5}{\ln 5 + \ln 7}
$$

4. **Calculate Numerically:**

Using approximate values:
- $$ \ln 5 \approx 1.6094 $$
- $$ \ln 7 \approx 1.9459 $$

$$
x \approx \frac{1.9459 - 1.6094}{1.6094 + 1.9459} = \frac{0.3365}{3.5553} \approx 0.0945
$$

**Answer:**  
$$ x = \dfrac{\ln 7 - \ln 5}{\ln 5 + \ln 7} \approx 0.0945 $$

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### **Problem (4): If $$ x^{2}+y^{2}=27xy $$, show that $$ \log \left(\frac{x - y}{5}\right) = \frac{1}{2}(\log x + \log y) $$.**

1. **Start with the Given Equation:**

$$
x^2 + y^2 = 27xy
$$

2. **Rearrange the Equation:**

$$
x^2 - 2xy + y^2 = 25xy
$$
$$
(x - y)^2 = 25xy
$$

3. **Divide Both Sides by 25:**

$$
\left( \frac{x - y}{5} \right)^2 = xy
$$

4. **Take the Square Root of Both Sides:**

$$
\frac{x - y}{5} = \sqrt{xy}
$$

5. **Apply the Logarithm:**

$$
\log\left(\frac{x - y}{5}\right) = \log(\sqrt{xy})
$$
$$
\log\left(\frac{x - y}{5}\right) = \frac{1}{2}(\log x + \log y)
$$

**Answer:**  
After simplifying, we have  
$$
\log\left(\frac{x - y}{5}\right) = \frac{1}{2}(\log x + \log y)
$$

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**Solutions:** 1. $$ \log 2512 \approx 3.4000 $$ 2. $$ \log \frac{x^2 y^3}{z^2} = 2 \log x + 3 \log y - 2 \log z $$ 3. $$ x = \frac{\ln 7 - \ln 5}{\ln 5 + \ln 7} \approx 0.0945 $$ 4. $$ \log \left(\frac{x - y}{5}\right) = \frac{1}{2}(\log x + \log y) $$

(1) Find the value of \( \log 2512 \) ? (2) Expand \( \log \frac{x 2 y 3}{z^{2}} \) (3) If \( 5^{x+1}=7^{1-x} \) Find \( X \) value. (4) If \( x^{2}+y^{2}=27 x y \), Show that \( \log \left(\frac{x-y}{5}\right)=\frac{1}{2}(\log x+\log y) \)

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