Question

The value of x, for which the 6th term in the expansion ${{\left\{ {{2}^{{{\log }_{2}}\sqrt{\left( {{9}^{x-1}}+7 \right)}}}+\dfrac{1}{{{2}^{(\frac{1}{5})}}{{\log }_{2}}({{3}^{x-1}}+1)} \right\}}^{7}}$ is 84, is equal to
A. 4
B. 3
C. 2
D. 0

Answer 1

Hint: In this question, we have given a term of some expansion whose value is also give. We solve these types of complex equations in parts. We divide the term into two parts, then we solve it one by one and get our desired answer.

Formula Used:
${{a}^{{{\log }_{b}}x}}={{x}^{{{\log }_{a}}b}}$

Complete step by step solution:
We have given 6th term in the expansion, which is ${{\left\{ {{2}^{{{\log }_{2}}\sqrt{\left( {{9}^{x-1}}+7 \right)}}}+\dfrac{1}{{{2}^{(\frac{1}{5})}}{{\log }_{2}}({{3}^{x-1}}+1)} \right\}}^{7}}$
And its value is 84.
We solve this equation in two parts
We know the formula of logarithm ${{a}^{{{\log }_{b}}x}}={{x}^{{{\log }_{a}}b}}$
We use the above equation to solve the equation ${{2}^{{{\log }_{2}}\sqrt{\left( {{9}^{x-1}}+7 \right)}}}$
${{2}^{{{\log }_{2}}\sqrt{\left( {{9}^{x-1}}+7 \right)}}}$= ${{\left( \sqrt{{{9}^{x-1}}+7} \right)}^{{{\log }_{2}}2}}$
By putting the value of ${{\log }_{2}}2$in the above equation, we get
${{\left( \sqrt{{{9}^{x-1}}+7} \right)}^{{{\log }_{2}}2}}$ = $\sqrt{{{9}^{x-1}}+7}$
Now we solve the equation ${{3}^{x}}=9,3$${{2}^{\frac{1}{5}{{\log }_{2}}\left( {{3}^{x-1}}+1 \right)}}$
Now the equation ${{2}^{\frac{1}{5}{{\log }_{2}}\left( {{3}^{x-1}}+1 \right)}}$ = ${{\left( {{3}^{x-1}}+1 \right)}^{\frac{1}{5}}}$
We know ${{T}_{r+1}}={}^{7}{{C}_{r}}{{\left( \sqrt{{{9}^{x-1}}+7} \right)}^{7-r}}\times {{\left( {{\left( {{3}^{x-1}}+1 \right)}^{\frac{-1}{5}}} \right)}^{r}}$
We know value of ${{T}_{6}} = 84$
${{T}_{6}}$ = ${}^{7}{{C}_{5}}{{\left( \sqrt{{{9}^{x-1}}+7} \right)}^{7-5}}\times {{\left( {{\left( {{3}^{x-1}}+1 \right)}^{\frac{-1}{5}}} \right)}^{5}}$
That is 84 = ${}^{7}{{C}_{5}}{{\left( \sqrt{{{9}^{x-1}}+7} \right)}^{7-5}}\times {{\left( {{\left( {{3}^{x-1}}+1 \right)}^{\frac{-1}{5}}} \right)}^{5}}$
Solving the above equation, we get
$21\times \dfrac{\left( {{3}^{2x-2}}+7 \right)}{{{3}^{x-1}}+1}$ = 84
Then we get $\dfrac{{{\left( {{3}^{x}} \right)}^{2}}}{9}+7=\dfrac{4}{3}\left( {{3}^{x}} \right)+4$
Now, suppose that ${{3}^{x}}=t$
We get the equation $\dfrac{{{t}^{2}}}{9}+7=\dfrac{4}{3}t+4$
By solving the above equation, we get a quadratic equation which is ${{t}^{2}}-12t+27=0$
Now, on solving the quadratic equation, we get (t-9) (t-3) = 0
Then, the value of t = 9,3
${{3}^{x}}=9,3$
Then the value of x = 2,1

Option ‘C’ is correct

Note: Whenever solving such types of questions, always write down the information provided in the question, and then use the standard formula of binomial expansion, as mentioned in the solution, i.e., $T_{r+1}={}^nC_ r a^{n−r}b^r$, to find the 6th term of the given expansion, and then equate it to find the values of x.