Question

If the fourth term in the expansion of \[{\left( {x + {x^{{{\log }_2}x}}} \right)^7}\] is 4480, then find the value of x, where \[x \in \mathbb{N}\] .
A.4
B.3
C.2
D.1

Answer 1

Hint: First write the fourth term of the expansion \[{\left( {x + {x^{{{\log }_2}x}}} \right)^7}\]. Then solve the quadratic equation for \[{\log _2}x\] and take the natural number value of x.

Formula used:
1. \[{\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{y^r}} \]
2. \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
3. \[\log MN = \log M + \log N\]
4. \[\log {a^b} = b\log a\]
5. \[{\log _a}a = 1\]

Complete step by step solution:
The fourth term of the given expression is \[{}^7{C_3}{x^{7 - 3}}{\left( {{x^{{{\log }_2}x}}} \right)^3}\]
That is, \[35{x^4}{\left( {{x^{{{\log }_2}x}}} \right)^3}\]
It is given that,
\[35{x^4}{\left( {{x^{{{\log }_2}x}}} \right)^3} = 4480\]
\[{x^4}{\left( {{x^{{{\log }_2}x}}} \right)^3} = 128\]
Take logarithm with respect to base 2 to both sides of the equation,
\[{\log _2}\left[ {{x^4}{{\left( {{x^{{{\log }_2}x}}} \right)}^3}} \right] = {\log _2}128\]
\[{\log _2}{x^4} + {\log _2}\left[ {{{\left( {{x^{{{\log }_2}x}}} \right)}^3}} \right] = {\log _2}{2^7}\]
\[{\log _2}{x^4} + {\log _2}\left[ {{{\left( {{x^{{{\log }_2}x}}} \right)}^3}} \right] = 7{\log _2}2\]
\[4{\log _2}x + \left( {3{{\log }_2}x} \right) \cdot {\log _2}x = 7\]
\[3{\left( {{{\log }_2}x} \right)^2} + 4{\log _2}x - 7 = 0\]
\[3{\left( {{{\log }_2}x} \right)^2} + 7{\log _2}x - 3{\log _2}x - 7 = 0\]
\[{\log _2}x\left( {3{{\log }_2}x + 7} \right) - 1\left( {3{{\log }_2}x + 7} \right) = 0\]
\[\left( {{{\log }_2}x - 1} \right)\left( {3{{\log }_2}x + 7} \right) = 0\]
But, \[\left( {3{{\log }_2}x + 7} \right) \ne 0\], as x is natural number.
So,
\[{\log _2}x - 1 = 0\]
\[x = {2^1}\]
Therefore, \[x = 2\] .

The correct answer is option (C).

Note: Sometime students write two values of x as the answer, but it is given that x is a natural number, therefore the fractional value cannot be taken as the answer.