Question

The constant term in the expansion of $\frac{3^x-2^x}{x^2}$ is

• A. ${\log }_{e}3$
• B. ${\log }_{e}6.{\log }_e\frac{3}{2}$
• C. $\frac{1}{2}{\log }_{e}6.{\log }_e\frac{3}{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{2}{\log }_{e}6.{\log }_e\frac{3}{2}$

$3^x=e^{\ln \left(3^x\right)}=1+x\ln \left(3\right)+\frac{x^2\ln (3{)}^2}{2!}+\frac{x^3\ln (3{)}^3}{3!}+\frac{x^4\ln (3{)}^4}{4!}...\infty$

$2^x=e^{\ln \left(2^x\right)}=1+x\ln \left(2\right)+\frac{x^2\ln (2{)}^2}{2!}+\frac{x^3\ln (2{)}^3}{3!}+\frac{x^4\ln (2{)}^4}{4!}...\infty$

Hence the term with $x^2$ in $3^x-2^x$ will be

$\frac{x^2\ln (3{)}^2}{2!}-\frac{x^2\ln (2{)}^2}{2!}$

$=\frac{x^2}{2}\left[\ln \right(3{)}^2-\ln \left(2{)}^2\right]$

$=\frac{x^2}{2}\left[\right(\ln \left(3\right)-\ln \left(2\right)\left)\ln \right(3\left)\ln \right(2\left)\right)]$

$=\frac{x^2}{2}\left[\ln \right(\frac{3}{2}\left)\ln \right(6\left)\right]$

Dividing by $x^2$, we get

$\ln \left(6\right)\ln \left(\frac{3}{2}\right)\frac{1}{2}$