Question

The value of x in the expression ${\left(x+x^{{\log }_{10}x}\right)}^5$, if the third term in the expansion is 1,000,000.
A. $10or{10}^{{\displaystyle \frac{-3}{2}}}$
B. $100or{10}^{{\displaystyle \frac{-3}{2}}}$
C. $10or{10}^{{\displaystyle \frac{-5}{2}}}$
D. None of these

Answer 1

Hint: In this question, first we substitute log${}_{10}$x=z so that the given expression becomes a binomial. Then we will write the general term of the binomial and use this to find the third term and equate it to ${10}^6$. Finally solve the equation to get the answer.

Complete step-by-step answer:
We have the expression${\left(x+x^{{\log }_{10}x}\right)}^5$.
In this question we need to find the value of x.
So let us substitute log${}_{10}$x=z.
Then the given expression will become
${\left(x+x^{{\log }_{10}x}\right)}^5$=$(x+x^z{)}^5$.
So, the expression now becomes a binomial.
We know that ${\left(x+y\right)}^n=\sum _{r=0}^n{}^n{C}_r{x}^{n-r}{y}^r$ and the general term is given by:
$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

Based on above expression, the binomial $(x+x^z{)}^5$ can be written as:
$(x+x^z{)}^5$=$\sum _{r=0}^5{}^5{C}_r{x}^{5-r}(x{)}^{zr}$
$T_{r+1}={}^5{C}_r{x}^{5-r}{x}^{zr}={}^5{C}_r{x}^{5-r+zr}$
So, for the third term, we will put r=2 in the above expression.
$T_3={}^5{C}_2{x}^{5-2+2z}={}^5{C}_2{x}^{3+2z}$
Now it is also given to us that $T_3={10}^6$.
So, on equating the third term to${10}^6$, we get:
${}^5{C}_2{x}^{3+2z}$=${10}^6$
And hence on simplification, we’ll have
${\displaystyle \frac{5!}{2!(5-2)!}}{x}^{3+2z}=$${10}^6$
$\Rightarrow {\displaystyle \frac{5\times 4\times 3!}{2\times 3!}}{x}^{3+2z}=$ ${10}^6$
And hence on further solving, we have:
$10x^{3+2z}={10}^6$
Now on taking 10 common from both the sides they both will cancel out each other and hence we have:
$\Rightarrow x^{3+2z}={10}^5$
Now on taking log both sides we have:
$$\Rightarrow \left(3+2z\right){\log }_{10}x=5{\log }_{10}10$$
$\Rightarrow \left(3+2z\right)z=5$
Now on doing the multiplication and then on simplifying we’ll have a quadratic equation and hence
$$\Rightarrow 2z^2+3z-5=0$$
Now since this is a quadratic equation and hence on doing the factorization, we have
( z – 1)( 2z + 5)=0
And hence z=$$\text{1,}{\displaystyle \frac{-5}{2}}$$
Therefore on putting the value of z we have,
$$\text{lo}{\text{g}}_{10}x=1\text{ or }{\displaystyle \frac{-5}{2}}$$
And hence we have x=$$10\text{ or 1}{\text{0}}^{{\displaystyle \frac{-5}{2}}}$$

So, the correct answer is “Option C”.

Note: In this type of question try to substitute log${}_{10}$x=z and hence on substituting and solving we’ll have a quadratic equation. You should know to write the general term of a binomial expression. Note that in the binomial expression,${\left(x+y\right)}^n=\sum _{r=0}^n{}^n{C}_r{x}^{n-r}{y}^r$, x, y $\in$R and ‘n’ must be a natural number.