Question

If $x>0$ and ${\log }_{2}x+{\log }_2\sqrt{x}+{\log }_2\sqrt[4]{4x}+{\log }_2\sqrt[8]{8x}+{\log }_2\sqrt[15]{15x}+....=4$, then $x=$______.

• A. $4$
• B. $2$
• C. $\pm 3$
• D. $\pm 48$

Answer 1

The correct option is A $4$

${\log }_{2}x+{\log }_2\sqrt{x}+{\log }_2\sqrt[4]{4x}+{\log }_2\sqrt[8]{8x}+\cdots \infty =4$

${\log }_2{x}^{1+\frac{1}{2}+\frac{1}{4}+\cdots \infty }=4$

${\log }_2{x}^2=4⟹x=4$