Question

The value of the expression $x={25}^{(\frac{1}{2}+{\log }_{1/5}27+{\log }_{125}81)}$ is

• A. $\frac{4}{81}$
• B. $1$
• C. a rational number
• D. an irrational number

Answer 1

The correct option is D an irrational number

let $y=\frac{1}{2}+{\log }_{\frac{1}{5}}27+{\log }_{125}81$
$\Rightarrow y=\frac{1}{2}-{\log }_5{3}^3+\frac{{\log }_5{3}^4}{3}[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b]$
$\Rightarrow y=\frac{1}{2}-3{\log }_{5}3+\frac{4}{3}{\log }_{5}3$
$\Rightarrow y={\log }_5\sqrt{5}-\frac{5}{3}{\log }_{5}3$
$\Rightarrow y={\log }_5\left(\frac{\sqrt{5}}{3^{5/3}}\right)[\because m\log a=\log a^m\text{\&}\log a-\log b=\log \frac{a}{b}]$
$x={25}^{(\frac{1}{2}+{\log }_{1/15}27+{\log }_{125}81)}$
$\Rightarrow x=5^{2{\log }_5\left(\sqrt{5}/3^{5/3}\right)}$
$\therefore x=\frac{5}{3^{10/3}}[\because a^{{\log }_{a}b}=b]$ is a irrational number.
Ans: D