Question

The value of $\log 3+\frac{(\log 3{)}^3}{3!}+\frac{(\log 3{)}^5}{5!}+\cdots \infty$

• A. $\frac{4}{3}$
• B. $\frac{5}{4}$
• C. $\frac{3}{4}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{4}{3}$

$sinhx=\frac{e^x-e^{-x}}{2}$
$=x+\frac{x^3}{3!}+\frac{x^5}{5!}...\infty$
Substituting $x=log3$
We get
$=log\left(3\right)+\frac{log(3{)}^3}{3!}+\frac{log(3{)}^5}{5!}...\infty$
$=sinh\left(log\right(3\left)\right)$
$=\frac{e^{log\left(3\right)}-e^{-log\left(3\right)}}{2}$
$=\frac{3-\frac{1}{3}}{2}$
$=\frac{8}{6}$
$=\frac{4}{3}$