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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.

**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 ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^{n - r}}{y^r}$ 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.

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