Question

The value of $1-\log 2+\frac{(\log 2{)}^2}{2!}-\frac{(\log 2{)}^3}{3!}+...$ is

• A. $\log 3$
• B. $\log 2$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option.

Find the value.

Given: $1-\log 2+\frac{(\log 2{)}^2}{2!}-\frac{(\log 2{)}^3}{3!}+...$

Formula ${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+.......$

In the question $\mathrm{x}=-\log 2$

Therefore

$$\begin{gathered} 1-\log 2+\frac{(\log 2{)}^2}{2!}-\frac{(\log 2{)}^3}{3!}+...={\mathrm{e}}^{-\log 2} \\ ={\mathrm{e}}^{\log 2^{-1}}\left[{\mathrm{e}}^{\mathrm{xlogy}}={\mathrm{e}}^{{\mathrm{logy}}^{\mathrm{x}}}\right] \\ =2^{-1}\left[{\mathrm{e}}^{\mathrm{loga}}=\mathrm{a}\right] \\ =\frac{1}{2} \end{gathered}$$

Hence $o\mathrm{ption}\left(\mathrm{C}\right)$ is the correct answer.