Question

lf $0<a<c,0<b<c$ then ${\int }_0^{\infty }\frac{a^x-b^x}{c^x}dx=$

• A. $\log \frac{b}{c}-\log \frac{a}{c}$
• B. $\frac{\log a-\log b}{\log c}$
• C. $\frac{1}{\log b/c}-\frac{1}{\log a/c}$
• D. $log\frac{a}{c}-log\frac{b}{c}$

Answer 1

The correct option is C $\frac{1}{\log b/c}-\frac{1}{\log a/c}$

${\int }_0^{\infty }(\frac{a^x}{c^x}-\frac{b^x}{c^x})dx$
$={\int }_0^{\infty }{\left(\frac{a}{c}\right)}^{x}dx-{\int }_0^{\infty }{\left(\frac{b}{c}\right)}^{x}dx$
$=\frac{(\frac{a}{c}{)}^x}{log\left(\frac{a}{c}\right)}{\int }_0^{\infty }-\frac{(\frac{b}{c}{)}^x}{log\left(\frac{b}{c}\right)}{\int }_0^{\infty }$
$=[1-\frac{1}{log\left(\frac{a}{c}\right)}]-[1-\frac{1}{log\left(\frac{b}{c}\right)}]$
$=\frac{1}{log\left(\frac{b}{c}\right)}-\frac{1}{log\left(\frac{a}{c}\right)}$