Find the value of xxx for which ∑n=1∞(log2x)n=2\sum_{n=1}^{\infty} (\log_2 x)^n = 2∑n=1∞(log2x)n=2.
x=2x = \sqrt{2}x=2
x=23x = \sqrt[3]{2}x=32
x=4x = 4x=4
x=2x = 2x=2