Question

Find the value of x in the expansion of\[{{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}\], if the third term in the expansion is 10,00,000. This question has multiple correct options.
A. \[{{10}^{-1}}\]
B. \[{{10}^{1}}\]
C. \[{{10}^{-5/2}}\]
D. \[{{10}^{5/2}}\]

Answer 1

Hint: Expand \[{{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}\]using binomial theorem formula that is given as: \[{{\left( a+b \right)}^{n}}=\sum\limits_{i=0}^{n}{\left( \begin{matrix}
  n \\
  i \\
\end{matrix} \right)}{{a}^{\left( n-i \right)}}{{b}^{i}}\], then equate the third term of this expansion with 10,00,000, to get the required value of x.

Complete step-by-step solution -
In the question, we have to find the value of x in the expansion of\[{{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}\], when the third term is given as 10,00,000.
So first, we will use the binomial theorem to find the expansion of \[{{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}\]. The binomial theorem for expansion\[{{\left( a+b \right)}^{n}}=\sum\limits_{i=0}^{n}{\left( \begin{matrix}
  n \\
  i \\
\end{matrix} \right)}{{a}^{\left( n-i \right)}}{{b}^{i}}\], where\[a=x,\;\;b={{x}^{{{\log }_{10}}\left( x \right)}}\]
So now the expansion form is:
\[\begin{align}
  & \Rightarrow {{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}=\sum\limits_{i=0}^{5}{\left( \begin{matrix}
  5 \\
  i \\
\end{matrix} \right)}{{x}^{\left( 5-i \right)}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{i}} \\
 & \Rightarrow {{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}=\dfrac{5!}{0!\left( 5-0 \right)!}{{x}^{5}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{0}}+\dfrac{5!}{1!\left( 5-1 \right)!}{{x}^{4}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{1}} \\
 & \,\,\,\,\,\,+\dfrac{5!}{2!\left( 5-2 \right)!}{{x}^{3}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{2}}+\dfrac{5!}{3!\left( 5-3 \right)!}{{x}^{2}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{3}}+\dfrac{5!}{4!\left( 5-4 \right)!}{{x}^{1}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{4}} \\
 & \,\,\,\,\,\,+\dfrac{5!}{5!\left( 5-5 \right)!}{{x}^{0}}{{\left( {{x}^{{{\log }_{10}}\left( x \right)}} \right)}^{5}} \\
 & \Rightarrow {{\left( x+{{x}^{lo{{g}_{10}}x}} \right)}^{5}}={{x}^{5}}+5{{x}^{4+{{\log }_{10}}\left( x \right)}}+10{{x}^{2{{\log }_{10}}\left( x \right)+3}}+10{{x}^{3{{\log }_{10}}\left( x \right)+2}}+5{{x}^{1+4{{\log }_{10}}\left( x \right)}}+{{x}^{5{{\log }_{10}}\left( x \right)}} \\
\end{align}\]
So, here the third term is \[10{{x}^{2{{\log }_{10}}\left( x \right)+3}}\]
It is also given that third term is 10,00,000. Hence, we will equate the above third term with this value to get the value of x. So, we get:
\[\begin{align}
  & \Rightarrow 10{{x}^{2{{\log }_{10}}\left( x \right)+3}}=1000000 \\
 & \Rightarrow 10{{e}^{\left( 2{{\log }_{10}}\left( x \right)+3 \right)\ln \left( x \right)}}=1000000\,\,\,\,\,\,\,\,\,\,\,\,\because {{k}^{f\left( x \right)}}={{e}^{f\left( x \right)\ln \left( k \right)}} \\
 & \Rightarrow {{e}^{\left( 2{{\log }_{10}}\left( x \right)+3 \right)\ln \left( x \right)}}=100000 \\
 & \Rightarrow \ln \left( {{e}^{\left( 2{{\log }_{10}}\left( x \right)+3 \right)\ln \left( x \right)}} \right)=\ln \left( 100000 \right) \\
 & \Rightarrow \left( 2{{\log }_{10}}\left( x \right)+3 \right)\ln \left( x \right)=\ln \left( 100000 \right) \\
 & \Rightarrow \left( 2{{\log }_{10}}\left( x \right)+3 \right)\dfrac{{{\log }_{10}}\left( x \right)}{{{\log }_{10}}\left( e \right)}=\ln \left( 100000 \right)\,\,\,\,\,\,\,\,\,\because {{\log }_{a}}\left( b \right)=\dfrac{{{\log }_{c}}\left( b \right)}{{{\log }_{c}}\left( a \right)} \\
 & \Rightarrow \dfrac{{{\log }_{10}}\left( x \right)\left( 2{{\log }_{10}}\left( x \right)+3 \right)}{{{\log }_{10}}\left( e \right)}=5\ln \left( 10 \right) \\
 & \Rightarrow {{\log }_{10}}\left( x \right)\left( 2{{\log }_{10}}\left( x \right)+3 \right)=5\,\,\,\,\,\,\,\,\,\,\because \ln \left( 10 \right){{\log }_{10}}\left( e \right)=1\,\,\,\, \\
\end{align}\]
Next take\[{{\log }_{10}}\left( x \right)=u\], we get the above equation as:
\[\begin{align}
  & \Rightarrow {{\log }_{10}}\left( x \right)\left( 2{{\log }_{10}}\left( x \right)+3 \right)=5 \\
 & \Rightarrow u\left( 2u+3 \right)=5 \\
 & \Rightarrow 2{{u}^{2}}+3u-5=0 \\
\end{align}\]
Next, this is the quadratic equation of the form \[a{{x}^{2}}+bx+c=0\], where x will be given by the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Now, in this equation \[2{{u}^{2}}+3u-5=0\] we have \[a=2,b=3,c=-5\]. So the value of u will be as follows:
\[\begin{align}
  & \Rightarrow 2{{u}^{2}}+3u-5=0 \\
 & \Rightarrow u=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\cdot \;2\left( -5 \right)}}{2\cdot \;2} \\
 & \Rightarrow u=\dfrac{-3\pm \sqrt{49}}{4} \\
 & \Rightarrow u=1,\;u=-\dfrac{5}{2} \\
\end{align}\]
Now, when \[u=1,\;\]we have:
\[\begin{align}
  & \Rightarrow u=1 \\
 & \Rightarrow {{\log }_{10}}\left( x \right)=1 \\
 & \Rightarrow \text{x=10} \\
\end{align}\]

Now, when \[u=-\dfrac{5}{2}\], we have:
\[\begin{align}
  & \Rightarrow u=-\dfrac{5}{2} \\
 & \Rightarrow {{\log }_{10}}\left( x \right)=-\dfrac{5}{2} \\
 & \Rightarrow x={{10}^{-\dfrac{5}{2}}} \\
\end{align}\]
So there are two possible values of x and they are \[\text{x=10}\] and \[x={{10}^{-\dfrac{5}{2}}}\].
Hence the correct answers are option B and option C.

Note: It is to be kept in mind that while using the substitution method in order to solve the equation, the value obtained is to be substituted back to the original variable to get the final answer.