Question

If the fourth term in the expansion of ${\left(x+x^{{\log }_{2}x}\right)}^7$ is $4480$, then the value of $x$ where $x\in N$ is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Step 1: Solve for the fourth term of the given expansion

Given expression is ${\left(x+x^{{\log }_{2}x}\right)}^7$ and the fourth term is $t_4=4480$

Applying binomial expansion, we know that ${\left(r+1\right)}^{th}$ term in the expansion of ${\left(a+b\right)}^n$is $t_{r+1}={}^{n}C_r{\left(a\right)}^{n-r}{\left(b\right)}^r$

$\therefore t_4={}^{7}C_3{\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3$

Step 2: Solve for the value of logarithm function

We need to evaluate ${\log }_2\left(x\right)$

$\begin{array}{crclc}& t_4& =& 4480& \\ \Rightarrow & {}^{7}C_3{\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3& =& 4480& \\ \Rightarrow & \frac{7!}{4!3!}{\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3& =& 4480& \left[\because {}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]\\ \Rightarrow & 35{\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3& =& 4480& \\ \Rightarrow & {\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3& =& \frac{4480}{35}& \\ \Rightarrow & {\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3& =& 128& \end{array}$

Apply ${\log }_2$ on both sides, we get,

$\begin{array}{crcll}& {\log }_2\left({\left(x\right)}^4{\left(x^{{\log }_2\left(x\right)}\right)}^3\right)& =& {\log }_2\left(128\right)& \\ \Rightarrow & {\log }_2\left(x^4\right)+{\log }_2{\left(x^{{\log }_2\left(x\right)}\right)}^3& =& {\log }_2{\left(2\right)}^7& \left[\because \log \left(mn\right)=\log \left(m\right)+\log \left(n\right)\right]\\ \Rightarrow & 4{\log }_2\left(x\right)+3{\log }_2\left(x\right){\log }_2\left(x\right)& =& 7{\log }_2\left(2\right)& \left[\because \log \left(m^n\right)=n\log \left(m\right)\right]\end{array}$

Step 3: Further simplify to obtain the value of ${\log }_2\left(x\right)$

Let ${\log }_2\left(x\right)=y$

$\begin{array}{crcll}\Rightarrow & 4y+3y^2& =& 7& \left[\because {\log }_2\left(2\right)=1\right]\\ \Rightarrow & 3y^2+4y-7& =& 0& \end{array}$

We know that, for quadratic equation $a{x}^2+bx+c=0;x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\begin{array}{crclc}\therefore & y& =& \frac{-4\pm \sqrt{{\left(4\right)}^2-4\left(3\right)\left(-7\right)}}{2\left(3\right)}& \\ & & =& \frac{-4\pm \sqrt{16+84}}{6}& \\ & & =& \frac{-4\pm \sqrt{100}}{6}& \\ & & =& \frac{-4\pm 10}{6}& \\ & & =& \frac{-4+10}{6},\frac{-4-10}{6}& \\ & & =& \frac{6}{6},\frac{-14}{6}& \\ \Rightarrow & y& =& 1,\frac{-7}{3}& \\ \Rightarrow & {\log }_2\left(x\right)& =& 1,\frac{-7}{3}& \end{array}$

Step 4: Solve for the value of $x$

$\begin{array}{crclc}& {\log }_2\left(x\right)& =& 1,\frac{-7}{3}& \\ \Rightarrow & x& =& 2^1,2^{\frac{-7}{3}}& \left[\because {\log }_m\left(x\right)=y\Rightarrow x=m^y\right]\end{array}$

Hence the correct option is option(C) i.e. $2$