Question

If $x={{\log }_{2}}\left( \sqrt{56+\sqrt{56+\sqrt{56+\sqrt{56+........\infty }}}} \right)$ , then which of the following statements hold good.
(A) \[x < 0\]
(B)\[0 < x < 2\]
(C)\[2 < x < 4\]
(D) \[3 < x < 4\]

Answer 1

Hint: To solve this question we need to have knowledge about exponential and logarithmic function. To solve the problem we are supposed to assume the term in the logarithmic function as an unknown variable. On changing we get a quadratic equation which will be solved, thus finding the value of $x$.

Complete step by step answer:
The question ask us to find the value of $x$ basically the range of $x$ for the question where $x={{\log }_{2}}\left( \sqrt{56+\sqrt{56+\sqrt{56+\sqrt{56+........\infty }}}} \right)$ is given.
The first step is to assume the term in the logarithmic function as an unknown variable.
We need to assume a variable $t$ which is equal to $\sqrt{56+t}$ . Which on mathematically is written as:
$\Rightarrow t=\sqrt{56+t}$
On squaring the both the term in LHS and in RHS we get:
$\Rightarrow {{t}^{2}}={{\left( \sqrt{56+t} \right)}^{2}}$
On expanding the above expression we get:
$\Rightarrow {{t}^{2}}=56+t$
On further calculating the term we get:
$\Rightarrow {{t}^{2}}-t-56=0$
On middle term factorisation, we will write the middle term as the sum of the two the terms, so the expression become:
$\Rightarrow {{t}^{2}}-8t+7t-56=0$
$\Rightarrow t\left( t-8 \right)+7\left( t-8 \right)=0$
$\Rightarrow \left( t-8 \right)\left( t+7 \right)=0$
So the value of $t=8,-7$
Now substituting the value of $t$ in the log function we get:
$\Rightarrow x={{\log }_{2}}8$ and $x={{\log }_{2}}\left( -7 \right)$
The log function cannot have negative value if the base is $2$, so we will only calculate only $x={{\log }_{2}}8$.
On calculating the log function we get:
$\Rightarrow x={{\log }_{2}}8$
$\Rightarrow x=3$
$\therefore $ If $x={{\log }_{2}}\left( \sqrt{56+\sqrt{56+\sqrt{56+\sqrt{56+........\infty }}}} \right)$ , then the option $C)2 < x < 4$ statements hold good.

Note: Do remember that the logarithmic function cannot have a negative value if the value has a positive base as given in the question is $2$. So in the given problem the term $x={{\log }_{2}}\left( -7 \right)$is not possible ${{2}^{x}}=-7$ as for any value of $x$ the value won’t give $-7$ as the answer.