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To solve the problem, let's go through each part step by step.

Given the function:
\[ y = (5 + 7x)^{2/x} \]

### Part (a): Logarithmic Differentiation
To use logarithmic differentiation, we take the natural logarithm of both sides:
\[ \ln y = \ln \left( (5 + 7x)^{2/x} \right) \]

Using the property of logarithms \(\ln(a^b) = b \ln(a)\):
\[ \ln y = \frac{2}{x} \ln(5 + 7x) \]

Now, differentiate both sides with respect to \( x \). For the left side, use the chain rule:
\[ \frac{d}{dx} (\ln y) = \frac{1}{y} \frac{dy}{dx} \]

For the right side, use the product rule:
\[ \frac{d}{dx} \left( \frac{2}{x} \ln(5 + 7x) \right) = \frac{d}{dx} \left( \frac{2}{x} \right) \ln(5 + 7x) + \frac{2}{x} \frac{d}{dx} (\ln(5 + 7x)) \]

Let's compute each term separately:
\[ \frac{d}{dx} \left( \frac{2}{x} \right) = -\frac{2}{x^2} \]

And,
\[ \frac{d}{dx} (\ln(5 + 7x)) = \frac{1}{5 + 7x} \cdot 7 = \frac{7}{5 + 7x} \]

Putting it all together:
\[ \frac{1}{y} \frac{dy}{dx} = -\frac{2}{x^2} \ln(5 + 7x) + \frac{2}{x} \cdot \frac{7}{5 + 7x} \]

Multiply both sides by \( y \) to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y \left( -\frac{2}{x^2} \ln(5 + 7x) + \frac{14}{x(5 + 7x)} \right) \]

Since \( y = (5 + 7x)^{2/x} \), we substitute back in:
\[ \frac{dy}{dx} = (5 + 7x)^{2/x} \left( -\frac{2}{x^2} \ln(5 + 7x) + \frac{14}{x(5 + 7x)} \right) \]

### Part (b): Slope of the Tangent Line at \( x = 1 \)
To find the slope at \( x = 1 \), we substitute \( x = 1 \) into the derivative:
\[ y = (5 + 7 \cdot 1)^{2/1} = 12^2 = 144 \]

Substitute \( x = 1 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=1} = 144 \left( -\frac{2}{1^2} \ln(5 + 7 \cdot 1) + \frac{14}{1(5 + 7 \cdot 1)} \right) \]

Calculate each term:
\[ \ln(12) \approx 2.4849 \]

\[ \frac{dy}{dx} \bigg|_{x=1} = 144 \left( -2 \cdot 2.4849 + \frac{14}{12} \right) = 144 \left( -4.9698 + 1.1667 \right) = 144 \left( -3.8031 \right) \approx -547.65 \]

### Part (c): Equation of the Tangent Line at \( x = 1 \)
The equation of the tangent line is given by:
\[ y - y_1 = m(x - x_1) \]

Where \( m \) is the slope at \( x = 1 \) and \( (x_1, y_1) \) is the point on the curve at \( x = 1 \).

We have \( x_1 = 1 \), \( y_1 = 144 \), and \( m \approx -547.65 \):
\[ y - 144 = -547.65(x - 1) \]

Simplify the equation:
\[ y = -547.65x + 547.65 + 144 \]

\[ y = -547.65x + 691.65 \]

Thus, the equation of the tangent line at \( x = 1 \) is:
\[ y \approx -547.65x + 691.65 \]

Do you want more details or have any questions? Here are some related questions you might find interesting:

1. How do you apply the chain rule in logarithmic differentiation?
2. Why is logarithmic differentiation useful for functions of the form \(a^b\)?
3. What is the product rule, and how is it applied in differentiation?
4. How do you interpret the derivative of a function at a given point?
5. What is the significance of the slope of a tangent line?
6. How do you find the equation of a tangent line given a point and a slope?
7. What are common mistakes to avoid when applying logarithmic differentiation?
8. How can you verify your derivative using alternative methods?

**Tip:** Always simplify logarithmic expressions before differentiating to make the process easier.

Learn how to solve the problem using logarithmic differentiation for the function (5 + 7x)^(2/x). Step-by-step explanation with detailed calculations and formulas. Suitable for college-level students and mathematics enthusiasts.

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