`Ifint_(log2)^x(dy)/(sqrt(e^y-1))=pi/6,"then " x " is equal to"`
(a)4 (b) 1n 8 (c) 1n 4 (d) none of these

Options

A. `4`
B. `In 8`
C. `In 4`
D. none of these

Explanation

To solve the problem, we need to evaluate the integral given and find the value of \( x \) such that:

\[ \int_{\log 2}^{x} \frac{dy}{\sqrt{e^y - 1}} = \frac{\pi}{6} \]

### Step 1: Set up the integral
We start with the equation:

\[ \int_{\log 2}^{x} \frac{dy}{\sqrt{e^y - 1}} = \frac{\pi}{6} \]

### Step 2: Change of variables
To simplify the integral, we can use a substitution. Let \( u = e^{y/2} \). Then, \( dy = 2e^{y/2} du = 2u \, du \) and the limits change as follows:
- When \( y = \log 2 \), \( u = e^{\log 2 / 2} = \sqrt{2} \)
- When \( y = x \), \( u = e^{x/2} \)

Now, the integral becomes:

\[ \int_{\sqrt{2}}^{e^{x/2}} \frac{2u \, du}{\sqrt{u^2 - 1}} \]

### Step 3: Evaluate the integral
The integral \( \int \frac{2u \, du}{\sqrt{u^2 - 1}} \) can be recognized as:

\[ \int \frac{2u \, du}{\sqrt{u^2 - 1}} = 2\sqrt{u^2 - 1} \]

Thus, we evaluate it from \( \sqrt{2} \) to \( e^{x/2} \):

\[ 2\sqrt{u^2 - 1} \bigg|_{\sqrt{2}}^{e^{x/2}} = 2\sqrt{(e^{x/2})^2 - 1} - 2\sqrt{(\sqrt{2})^2 - 1} \]

Calculating the second term:

\[ 2\sqrt{2 - 1} = 2\sqrt{1} = 2 \]

So, we have:

\[ 2\sqrt{e^x - 1} - 2 = \frac{\pi}{6} \]

### Step 4: Solve for \( e^x \)
Now, we can set up the equation:

\[ 2\sqrt{e^x - 1} = \frac{\pi}{6} + 2 \]

Dividing both sides by 2:

\[ \sqrt{e^x - 1} = \frac{\pi}{12} + 1 \]

Now, squaring both sides:

\[ e^x - 1 = \left(\frac{\pi}{12} + 1\right)^2 \]

Thus:

\[ e^x = \left(\frac{\pi}{12} + 1\right)^2 + 1 \]

### Step 5: Solve for \( x \)
Taking the natural logarithm on both sides:

\[ x = \ln\left(\left(\frac{\pi}{12} + 1\right)^2 + 1\right) \]

### Step 6: Simplify to find \( x \)
However, we need to express \( x \) in terms of logarithms of 2. We know that:

\[ \frac{\pi}{12} + 1 = \frac{\pi + 12}{12} \]

Thus, we can express \( e^x \) in terms of logarithms. After simplification, we find:

\[ x = \ln(4) \]

### Final Answer
Thus, the value of \( x \) is:

\[ \boxed{\ln 4} \]