Question

Find the value of ‘x’ where the fourth term in the expansion,
 $ {\left( {{5^{\dfrac{2}{5}{{\log }_5}\sqrt {{4^x} + 44} }} + \dfrac{1}{{{5^{{{\log }_5}\sqrt[3]{{{2^{x - 1}} + 7}}}}}}} \right)^8} $ is 336.

Answer 1

Hint: The above question revolves around the concept of binomial theorem and we observe that the expression is raised to the power 8. Therefore the expansion will have nine terms in total out of which we are given with the fourth term which also happens to be the middle term of the expansion.
Formula used: some properties of logarithm used are:
 $
  {a^{{{\log }_a}x}} = x \\
  {\log _{{a^m}}}{x^n} = \dfrac{n}{m}{\log _a}x \\
 $

Complete step-by-step answer:
We must know that the general term in a binomial expansion is given by
 $ \Rightarrow {T_{r + 1}}{ = ^n}{C_r}{x^{n - r}}{a^r} $
We have been given a fourth term therefore the value of r=3 and n=8 which is the power of the expansion. Hence fourth term from the expansion is
 $ \Rightarrow {T_4}{ = ^8}{C_3}.{x^{8 - 3}}.{a^3}{ = ^8}{C_3}.{x^5}.{a^3} $
Where a binomial expression is of the form $ {(x + a)^n} $ and we can compare the value of x and a from the given expression. Hence fourth term can be written as,
 $ \Rightarrow {T_4}{ = ^8}{C_3}{({5^{\dfrac{2}{5}{{\log }_5}\sqrt {{4^x} + 44} }})^5}.{(\dfrac{1}{{{5^{{{\log }_5}\sqrt[3]{{{2^{x - 1}} + 7}}}}}})^3} $
Let us expand x and a individually and then put the derived value in the above equation.
The logarithmic term in x can be evaluated by using the formula listed as,
 $ \Rightarrow {5^{\dfrac{2}{5}{{\log }_5}\sqrt {{4^x} + 44} }} = {5^{{{\log }_5}{{\left( {{4^x} + 44} \right)}^{\dfrac{1}{2} \times \dfrac{2}{5}}}}} = {\left( {{4^x} + 44} \right)^{\dfrac{1}{5}}} $
The logarithmic value in a can also be evaluated similarly,
 $ \Rightarrow \dfrac{1}{{{5^{{{\log }_5}\sqrt[3]{{{2^{x - 1}} + 7}}}}}} = \dfrac{1}{{{5^{{{\log }_5}{{\left( {{2^{x - 1}} + 7} \right)}^{\dfrac{1}{3}}}}}}} = \dfrac{1}{{^{{{\left( {{2^{x - 1}} + 7} \right)}^{\dfrac{1}{3}}}}}} $
Now let us put these calculated values of a and x in the expression of fourth term,
 $
\Rightarrow {T_4}{ = ^8}{C_3}{({\left( {{4^x} + 44} \right)^{\dfrac{1}{5}}})^5}.{(\dfrac{1}{{{{({2^{x - 1}} + 7)}^{\dfrac{1}{3}}}}})^3} \\
  {T_4}{ = ^8}{C_3}.\left( {{4^x} + 44} \right).(\dfrac{1}{{({2^{x - 1}} + 7)}}) \\
 $
It can be further expanded,
 $
\Rightarrow {T_4} = \dfrac{{\left| \!{\underline {\,
  8 \,}} \right. }}{{\left| \!{\underline {\,
  3 \,}} \right. .\left| \!{\underline {\,
  5 \,}} \right. }}.\left( {{4^x} + 44} \right).(\dfrac{1}{{({2^{x - 1}} + 7)}}) \\
   = \dfrac{{8 \times 7 \times 6}}{6}.\left( {{4^x} + 44} \right).(\dfrac{1}{{({2^{x - 1}} + 7)}}) = 56.\dfrac{{({2^{2x}} + 44)}}{{{2^{x - 1}} + 7}} \\
  336 = 56.\dfrac{{2.({2^{2x}} + 44)}}{{{2^x} + 14}} = 56.\dfrac{{2.({{({2^x})}^2} + 44)}}{{{2^x} + 14}} \;
 $
Let us substitute $ {2^x} = y $ in the given equation,
 $
  336 = 56.\dfrac{{2.({{(y)}^2} + 44)}}{{y + 14}} \\
  \dfrac{{336}}{{112}} = 3 = \dfrac{{{y^2} + 44}}{{y + 14}} \\
  3(y + 14) = {y^2} + 44 \\
  3y + 42 = {y^2} + 44 \\
  {y^2} - 3y + 2 = 0 \;
 $
Finding the value of y by factorization,
(y-1)(y-2)=0
Y=1 and y=2
 $
\Rightarrow y = {2^x} \\
\Rightarrow y = 1,2 \\
\Rightarrow {2^x} = 1,x = 0 \\
\Rightarrow {2^x} = 2,x = 1 \;
 $
Hence we have two values for x as 0 and 1.
So, the correct answer is “0 AND 1”.

Note: In binomial expression whenever we have to find any term from the end, find which term it is from the beginning then evaluate in the way similar like above using the formula mentioned above.