If x = _2 ( 56 + 56 + 56 + .... + ) then - Vidyadip

MCQ

If $x = {\log _2}\left( {\sqrt {56 + \sqrt {56 + \sqrt {56 + .... + \infty } } } } \right)$ then

A
$x < 0$
B
$0 < x < 2$
✓
$2 < x < 4$
D
$3 < x < 4$

✓

Answer

Correct option: C.

$2 < x < 4$

c
$x=\log _2(\sqrt{56+\sqrt{56+\ldots \infty}})$

het $t=\sqrt{56+t}$

$t^2-t-56=0$

$(t+7)(t-8)=0$

$t=8,-7 \quad$ [log can't be negative]

$\therefore x=\log _2 8$

$\therefore x=3$ (c)

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