Question

If $f\left(x\right)={\left(\frac{5}{4}\right)}^x,then{\int }_0^{\infty }\left[\frac{2}{f\left(x\right)}\right]dx$, where [.] denotes greatest integer function, equals

• A. ${\log }_{\frac{5}{4}}\left(\frac{1}{2}\right)$
• B. ${\log }_{\frac{3}{4}}\left(\frac{1}{2}\right)$
• C. ${\log }_{\frac{4}{5}}\left(2\right)$
• D. ${\log }_{\frac{4}{5}}\left(\frac{1}{2}\right)$

Answer 1

The correct option is A ${\log }_{\frac{5}{4}}\left(\frac{1}{2}\right)$

Given,

$f\left(x\right)={\left(\frac{5}{4}\right)}^x$

${\int }_0^{\infty }\left(\frac{2}{f\left(x\right)}\right)dx$

${\int }_0^{\infty }\left(\frac{2}{{\left(\frac{5}{4}\right)}^x}\right)dx$

Now,

$\int \frac{2}{{\left(\frac{5}{4}\right)}^x}dx$

$=2\cdot \int {\left(\frac{4}{5}\right)}^{x}dx$

$=2\cdot \frac{{\left(\frac{4}{5}\right)}^x}{\ln \left(\frac{4}{5}\right)}$

$=\frac{2^{2x+1}}{\ln \left(\frac{4}{5}\right)\cdot 5^x}$

$=\frac{2^{2x+1}}{\ln \left(\frac{4}{5}\right)\cdot 5^x}+C$

${\int }_0^{\infty }\frac{2}{{\left(\frac{5}{4}\right)}^x}dx={\left[\frac{2^{2x+1}}{\ln \left(\frac{4}{5}\right)\cdot 5^x}\right]}_0^{\infty }$

$=0-\frac{2}{\ln \left(\frac{4}{5}\right)}$

$=-\frac{2}{\ln \left(\frac{4}{5}\right)}$

$={\log }_{\frac{5}{4}}\left(\frac{1}{2}\right)$