Question

If the fourth term in the binomial expansion of \[{{\left( \dfrac{2}{x}+{{x}^{{{\log }_{8}}x}} \right)}^{6}}\left( x>0 \right)\] is \[20\times {{8}^{7}},\] then the value of x is?
\[\left( a \right)8\]
\[\left( b \right){{8}^{-1}}\]
\[\left( c \right){{8}^{-2}}\]
\[\left( d \right){{8}^{3}}\]

Answer 1

Hint: The general \[{{\left( r+1 \right)}^{th}}\] term in any binomial is given as \[{{T}_{r+1}}={{\text{ }}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}.\] So we will use this at first and then we use log properties like \[\log {{a}^{n}}=n\log a,\] if \[{{\log }_{a}}b=c\] then we have \[b={{a}^{c}}.\] Further, we will substitute \[t={{\log }_{2}}x\] to simplify our equation. We will get a quadratic equation and will simplify it using the middle term split method. Then out of all the answers, we will choose an option available to us.

Complete step-by-step answer:
We are given that the fourth term of the binomial expansion \[{{\left( \dfrac{2}{x}+{{x}^{{{\log }_{8}}x}} \right)}^{6}}\] is \[20\times {{8}^{7}}.\] We are asked to find x. We know that for any binomial expansion is \[{{\left( a+b \right)}^{n}}.\]
The \[{{\left( r+1 \right)}^{th}}\] term \[{{T}_{r+1}}\] is given as \[{{T}_{r+1}}={{\text{ }}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}.\]
So, for \[{{\left( \dfrac{2}{x}+{{x}^{{{\log }_{8}}x}} \right)}^{6}}\] we have \[a=\dfrac{2}{x},b={{x}^{{{\log }_{8}}x}},n=6.\] So, the fourth term will be given as, \[{{T}_{4}}={{T}_{3+1}}.\] So,
\[{{T}_{3+1}}={{\text{ }}^{6}}{{C}_{3}}{{\left( \dfrac{2}{x} \right)}^{3}}{{\left( {{x}^{{{\log }_{8}}x}} \right)}^{3}}\]
Also, we have,
\[{{T}_{4}}=20\times {{8}^{7}}\]
So, comparing both, we get,
\[{{\text{ }}^{6}}{{C}_{3}}{{\left( \dfrac{2}{x} \right)}^{3}}{{\left( {{x}^{{{\log }_{8}}x}} \right)}^{3}}=20\times {{8}^{7}}\]
Simplifying further, we get,
\[\Rightarrow \dfrac{160}{{{x}^{3}}}\times {{\left( {{x}^{{{\log }_{8}}x}} \right)}^{3}}=20\times {{8}^{7}}\]
Now, \[{{\left( {{x}^{{{\log }_{8}}x}} \right)}^{3}}={{x}^{3{{\log }_{8}}x}}.\] So, we get,
\[\Rightarrow \dfrac{160}{{{x}^{3}}}{{x}^{3{{\log }_{8}}x}}=20\times {{8}^{7}}\]
Cancelling the like terms, we get,
\[\Rightarrow \dfrac{{{x}^{3{{\log }_{8}}x}}}{{{x}^{3}}}={{8}^{6}}\]
As, \[\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}},\] we get,
\[\Rightarrow {{x}^{3{{\log }_{8}}x-3}}={{8}^{6}}\]
Now, we know that, \[8={{2}^{3}}.\] So, we get,
\[\Rightarrow {{8}^{6}}={{\left( {{2}^{3}} \right)}^{6}}={{2}^{18}}\]
\[\Rightarrow {{x}^{3{{\log }_{{{2}^{3}}}}x-3}}={{2}^{18}}\]
Simplifying, we get,
\[\Rightarrow {{x}^{{{\log }_{2}}x-3}}={{2}^{18}}\]
Appling \[{{\log }_{2}}\] on both the sides, we get,
\[\Rightarrow {{\log }_{2}}\left( {{x}^{{{\log }_{2}}x-3}} \right)={{\log }_{2}}{{2}^{18}}\]
We know that, \[\log {{a}^{n}}=n\log a,\] so, we get,
\[\Rightarrow \left( {{\log }_{2}}x-3 \right){{\log }_{2}}x=18{{\log }_{2}}2\]
We know that, \[{{\log }_{a}}\left( a \right)=1.\] So, \[{{\log }_{2}}\left( 2 \right)=1.\]
\[\Rightarrow \left( {{\log }_{2}}x-3 \right){{\log }_{2}}x=18\]
Taking \[{{\log }_{2}}x=t\] to simplify, we get,
\[\Rightarrow \left( t-3 \right)t=18\]
\[\Rightarrow {{t}^{2}}-3t-18=0\]
Solving, we get,
\[\Rightarrow \left( t-6 \right)\left( t+3 \right)=0\]
So, we get, t = 6 and t = – 3.
(i) If t = 6
This means, \[{{\log }_{2}}x=6\]
We know that \[{{\log }_{a}}b=c.\] From this, we get,
\[\Rightarrow b={{a}^{c}}\]
So, \[{{\log }_{2}}x=6.\]
\[\Rightarrow x={{2}^{6}}\]
\[\Rightarrow x={{\left( {{2}^{3}} \right)}^{6}}\]
\[\Rightarrow x={{8}^{2}}\]
(ii) If t = – 3
This means, \[{{\log }_{2}}x=-3\]
We know that \[{{\log }_{a}}b=c.\] From this, we get,
\[\Rightarrow b={{a}^{c}}\]
So, \[{{\log }_{2}}x=-3.\]
\[\Rightarrow x={{2}^{-3}}\]
\[\Rightarrow x={{8}^{-1}}\]
So, option (b) is matching.

So, the correct answer is “Option b”.

Note: Remember that we simplified our quadratic equation using the middle term split method.
\[\Rightarrow {{t}^{2}}-3t-18=0\]
\[\Rightarrow {{t}^{2}}-6t+3t-18=0\]
Taking the common, we get,
\[\Rightarrow t\left( t-6 \right)+3\left( t-6 \right)=0\]
\[\Rightarrow \left( t+3 \right)\left( t-6 \right)=0\]
So,
\[\Rightarrow t+3=0;t-6=0\]
\[\Rightarrow t=-3;t=6\]
Also, as, \[{{\log }_{a}}b=c,\] we get, \[b={{a}^{c}}.\]
Then applying it at \[{{\log }_{2}}x=6,\] we have, a = 2, b = x and c = 6. So, \[b={{a}^{c}}\] means \[x={{2}^{6}}={{\left( {{2}^{3}} \right)}^{2}}={{8}^{2}}.\]