If x = log _2 left( sqrt {56 + sqrt {56 + sqr | vedclass

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(C) Let $t = \sqrt{56 + \sqrt{56 + \sqrt{56 + \dots \infty}}}$.Then $t = \sqrt{56 + t}$.Squaring both sides,we get $t^2 = 56 + t$,which implies $t^2 -

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$2 < x < 4$

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(C) Let $t = \sqrt{56 + \sqrt{56 + \sqrt{56 + \dots \infty}}}$.Then $t = \sqrt{56 + t}$.Squaring both sides,we get $t^2 = 56 + t$,which implies $t^2 - t - 56 = 0$.Factoring the quadratic equation,we get $(t - 8)(t + 7) = 0$.Since $t$ represents a square root,$t$ must be positive,so $t = 8$.Now,substitute $t$ into the expression for $x$: $x = \log _2(8)$.Since $8 = 2^3$,we have $x = \log _2(2^3) = 3$.Since $x = 3$,it satisfies the condition $2 < x < 4$.

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

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