Let \(x\) = \(2^{\frac{1}{4}}\) x  \(4^{\frac{1}{8}}\) x \(8^{\frac{1}{16}}\) x \(16^{\frac{1}{32}}\)......∞

∴ log \(x\) = \(\frac{1}{4}\) log 2 + \(\frac{1}{8}\) log 4 + \(\frac{1}{16}\) log 8 + \(\frac{1}{32}\) log 16 + ..........∞

=   \(\frac{1}{4}\) log 2 + \(\frac{1}{8}\) log 2² + \(\frac{1}{16}\) log 2³ + \(\frac{1}{32}\) log 2⁴ + ..........∞

=   \(\frac{1}{4}\) log 2 + \(\frac{2}{8}\) log 2 + \(\frac{3}{16}\) log 2 + \(\frac{4}{32}\) log 2 + ..........∞

= \(\big(\)\(\frac{1}{4}\) + \(\frac{2}{8}\) + \(\frac{3}{16}\) + \(\frac{4}{32}\) + .........∞\(\big)\) log 2               .....(i)

Now, \(\frac{1}{4}\) + \(\frac{2}{8}\) + \(\frac{3}{16}\) + \(\frac{4}{32}\) + ....∞ is an Arithmetico-Geometric series.

Let  S =   \(\frac{1}{4}\) + \(\frac{2}{8}\) + \(\frac{3}{16}\) + \(\frac{4}{32}\) + .......∞             .....(ii)

\(\frac{1}{2}\)S = \(\frac{1}{8}\) +\(\frac{2}{16}\) + \(\frac{3}{32}\) + .......∞                ........(iii)

On subtracting eqn (iii) from eqn (ii), we get

 \(\frac{1}{2}\)S = \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + ........∞ = \(\frac{\frac{1}{4}}{1-\frac{1}{2}}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) = \(\frac{1}{2}\) ⇒ S = 1

∴ log \(x\) = 1 × log 2                     (From (i)) 

⇒ log \(x\) = log 2 ⇒ \(x\) = 2.