Question

The value of \[S=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] , then S is equal to
A.\[{{e}^{2}}\]
B.\[{{a}^{2}}\]
C.\[{{2}^{a}}\]
D.\[{{2}^{x}}\]

Answer 1

Hint: According to the question, we have the series, \[S=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] . We know the expansion of \[{{e}^{x}}\] , \[{{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{2}}}{2!}+............\infty \] . Replace x by \[x{{\log }_{e}}2\] in the expansion of \[{{e}^{x}}\] . Using the formula \[m\log n=\log {{n}^{m}}\] , solve \[x{{\log }_{e}}2\] . Now, we also know the formula, \[{{e}^{{{\log }_{e}}x}}=x\] . Use this formula and solve it further.

Complete step-by-step answer:
According to the question, it is given that our series is,
\[S=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] ……………………….(1)
Here, we have terms like factorials in the above equation. So, we have to think about some series which includes terms like factorials and also the summation should be known to us.
Here, we can think of the expansion of \[{{e}^{x}}\] . The expansion of \[{{e}^{x}}\] is
\[{{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{2}}}{2!}+............\infty \] ………………….(2)
Now, replacing x by \[x{{\log }_{e}}2\] in equation (2), we get
\[{{e}^{x{{\log }_{e}}2}}=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] ………………..(2)
We know the formula that, \[m\log n=\log {{n}^{m}}\] .
Now, using this formula in equation (2), we get
\[{{e}^{x{{\log }_{e}}2}}=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \]
\[\Rightarrow {{e}^{{{\log }_{e}}{{2}^{x}}}}=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] …………………..(3)
We also know the formula, \[{{e}^{{{\log }_{e}}x}}=x\] .
Using this formula in equation (3), we get
\[\Rightarrow {{e}^{{{\log }_{e}}{{2}^{x}}}}=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \]
\[\Rightarrow {{2}^{x}}=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] …………………..(4)
From equation (1), we have \[S=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \] .
On comparing equation (1) and equation (4), we get
\[S={{2}^{x}}\] .
So, the value of x is \[{{2}^{x}}\] .
Hence, the correct option is option (D).

Note: For solving this type of questions, one must remember the expansion of \[{{e}^{x}}\] . In this question, one may think to multiply S by \[x{{\log }_{e}}2\] and then subtract it from S.
\[S=1+\dfrac{x{{\log }_{e}}2}{1!}+\dfrac{{{x}^{2}}{{({{\log }_{e}}2)}^{2}}}{2!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{3!}+................\infty \]
\[S\left( x{{\log }_{e}}2 \right)=x{{\log }_{e}}2+\dfrac{{{(x{{\log }_{e}}2)}^{2}}}{1!}+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{2!}+\dfrac{{{x}^{4}}{{({{\log }_{e}}2)}^{4}}}{3!}+................\infty \]
After subtracting we get,
\[S\left( 1-x{{\log }_{e}}2 \right)=x{{\log }_{e}}2+\dfrac{{{(x{{\log }_{e}}2)}^{2}}}{1}\left( \dfrac{1}{2!}-\dfrac{1}{1!} \right)+\dfrac{{{x}^{3}}{{({{\log }_{e}}2)}^{3}}}{1}\left( \dfrac{1}{3!}-\dfrac{1}{2!} \right)+\dfrac{{{x}^{4}}{{({{\log }_{e}}2)}^{4}}}{1}\left( \dfrac{1}{4!}-\dfrac{1}{3!} \right)+................\infty \]
In the above equation, we can see that our approach made this summation even more complex to solve. So, we don’t have to approach this question by this method.