We are given the integral equation: \[ \int_{\log 2}^{x} \frac{1}{\sqrt{e^t - 1}} \, dt = \frac{\pi}{6}. \] Step 1: Substituting \( e^t = u \)
Let: \[ u = e^t \Rightarrow du = e^t dt. \] Thus, \[ dt = \frac{du}{u}. \] Rewriting the integral in terms of \( u \): \[ \int_{\log 2}^{x} \frac{dt}{\sqrt{e^t - 1}}. \] Using \( u = e^t \), we transform the integral: \[ \int_{e^{\log 2}}^{e^x} \frac{\frac{du}{u}}{\sqrt{u - 1}}. \] Since \( e^{\log 2} = 2 \), the limits change from \( 2 \) to \( e^x \): \[ \int_{2}^{e^x} \frac{du}{u \sqrt{u - 1}}. \] Step 2: Substituting \( u - 1 = v^2 \) Let: \[ u - 1 = v^2 \Rightarrow du = 2v dv. \] Substituting in the integral: \[ \int_{2}^{e^x} \frac{2v dv}{(v^2 + 1) \sqrt{v^2}}. \] Simplifying, \[ \int_{v_0}^{v_1} \frac{2v dv}{(v^2 + 1) v} = \int_{v_0}^{v_1} \frac{2 dv}{v^2 + 1}. \] Since \( \int \frac{2 dx}{x^2 + 1} = 2 \tan^{-1} x \), the integral evaluates to: \[ 2 \tan^{-1} v \Big|_{v_0}^{v_1}. \] For limits,
\( v_0 = \sqrt{2 - 1} = 1 \).
\( v_1 = \sqrt{e^x - 1} \).
Thus, we get: \[ 2 \left(\tan^{-1} \sqrt{e^x - 1} - \tan^{-1} 1\right) = \frac{\pi}{6}. \] Since \( \tan^{-1} 1 = \frac{\pi}{4} \), we simplify: \[ 2 \tan^{-1} \sqrt{e^x - 1} - \frac{\pi}{2} = \frac{\pi}{6}. \] \[ 2 \tan^{-1} \sqrt{e^x - 1} = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3}. \] \[ \tan^{-1} \sqrt{e^x - 1} = \frac{\pi}{3}. \] Step 3: Solving for \( x \)
Taking tangent on both sides: \[ \sqrt{e^x - 1} = \tan \frac{\pi}{3}. \] Since \( \tan \frac{\pi}{3} = \sqrt{3} \), we square both sides: \[ e^x - 1 = 3. \] \[ e^x = 4. \] \[ x = \log 4. \] Since \( \log 4 = 2 \log 2 \),
we conclude: \( 2 \log 2 \)