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# 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. $10\;or\;{10^{\dfrac{{ - 3}}{2}}}$B. $100\;or\;{10^{\dfrac{{ - 3}}{2}}}$C. $10\;or\;{10^{\dfrac{{ - 5}}{2}}}$D. None of these

Last updated date: 14th Jun 2024
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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.

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 ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^{n - r}}{y^r}$ 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:
${(x + {x^z})^5}$=$\sum\limits_{r = 0}^5 {{}^5{C_r}} {x^{5 - r}}{(x)^{zr}}$
${T_{r + 1}} = {}^5{C_r}{x^{5 - r}}{x^{zr}} = {}^5{C_r}{x^{5 - r + zr}}$
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
$\dfrac{{5!}}{{2!(5 - 2)!}}{x^{3 + 2z}} =$${10^6}$
$\Rightarrow \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}}{x^{3 + 2z}} =$ ${10^6}$
And hence on further solving, we have:
$10{x^{3 + 2z}} = {10^6}$
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
$\Rightarrow 2{z^2} + 3z - 5 = 0$
Now since this is a quadratic equation and hence on doing the factorization, we have
( z – 1)( 2z + 5)=0
And hence z=${\text{1,}}\dfrac{{ - 5}}{2}$
Therefore on putting the value of z we have,
${\text{lo}}{{\text{g}}_{10}}x = 1{\text{ or }}\dfrac{{ - 5}}{2}$
And hence we have x=$10{\text{ or 1}}{{\text{0}}^{\dfrac{{ - 5}}{2}}}$

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,${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^{n - r}}{y^r}$, x, y $\in$R and ‘n’ must be a natural number.