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Question: The value of x in the expression \({\left( {x + {x^{{{\log }_{10}}x}}} \right)^5}\), if the third te...

The value of x in the expression (x+xlog⁡10x)5{\left( {x + {x^{{{\log }_{10}}x}}} \right)^5}, if the third term in the expansion is 1,000,000.
A. 10  or  10−3210\;or\;{10^{\dfrac{{ - 3}}{2}}}
B. 100  or  10−32100\;or\;{10^{\dfrac{{ - 3}}{2}}}
C. 10  or  10−5210\;or\;{10^{\dfrac{{ - 5}}{2}}}
D. None of these

Explanation

Solution

In this question, first we substitute log10_{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 106{10^6}. Finally solve the equation to get the answer.

Complete step-by-step answer:
We have the expression(x+xlog⁡10x)5{\left( {x + {x^{{{\log }_{10}}x}}} \right)^5}.
In this question we need to find the value of x.
So let us substitute log10_{10}x=z.
Then the given expression will become
(x+xlog⁡10x)5{\left( {x + {x^{{{\log }_{10}}x}}} \right)^5}=(x+xz)5{(x + {x^z})^5}.
So, the expression now becomes a binomial.
We know that (x+y)n=∑r=0nnCrxn−ryr{\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:
Tr+1=nCrxn−ryr{T_{r + 1}} = {}^n{C_r}{x^{n - r}}{y^r}

Based on above expression, the binomial (x+xz)5{(x + {x^z})^5} can be written as:
(x+xz)5{(x + {x^z})^5}=∑r=055Crx5−r(x)zr\sum\limits_{r = 0}^5 {{}^5{C_r}} {x^{5 - r}}{(x)^{zr}}
Tr+1=5Crx5−rxzr=5Crx5−r+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.
T3=5C2x5−2+2z=5C2x3+2z{T_3} = {}^5{C_2}{x^{5 - 2 + 2z}} = {}^5{C_2}{x^{3 + 2z}}
Now it is also given to us that T3=106{T_3} = {10^6}.
So, on equating the third term to106{10^6}, we get:
5C2x3+2z{}^5{C_2}{x^{3 + 2z}}=106{10^6}
And hence on simplification, we’ll have
5!2!(5−2)!x3+2z=\dfrac{{5!}}{{2!(5 - 2)!}}{x^{3 + 2z}} = 106{10^6}
⇒5×4×3!2×3!x3+2z=\Rightarrow \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}}{x^{3 + 2z}} = 106{10^6}
And hence on further solving, we have:
10x3+2z=10610{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:
⇒x3+2z=105\Rightarrow {x^{3 + 2z}} = {10^5}
Now on taking log both sides we have:
⇒(3+2z)log⁡10x=5log⁡1010\Rightarrow \left( {3 + 2z} \right){\log _{10}}x = 5{\log _{10}}10
⇒(3+2z)z=5\Rightarrow \left( {3 + 2z} \right)z = 5
Now on doing the multiplication and then on simplifying we’ll have a quadratic equation and hence
⇒2z2+3z−5=0\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=1,−52{\text{1,}}\dfrac{{ - 5}}{2}
Therefore on putting the value of z we have,
log10x=1 or −52{\text{lo}}{{\text{g}}_{10}}x = 1{\text{ or }}\dfrac{{ - 5}}{2}
And hence we have x=10 or 10−5210{\text{ or 1}}{{\text{0}}^{\dfrac{{ - 5}}{2}}}

So, the correct answer is “Option C”.

Note: In this type of question try to substitute log10_{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,(x+y)n=∑r=0nnCrxn−ryr{\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^{n - r}}{y^r}, x, y ∈\inR and ‘n’ must be a natural number.