Question

If $(2.3{)}^x=(0.23{)}^y=1000$ find $\frac{1}{x}-\frac{1}{y}$

Answer 1

Given $(2.3{)}^x=(0.23{)}^y=1000$
Consider $(2.3{)}^x=1000$
taking ${\log }_{2.3}$ on both sides of above equation
${\log }_{2.3}1000=x$
$\frac{1}{{\log }_{1000}2.3}=x$ (Since, ${\log }_{m}n=\frac{1}{{\log }_{n}m}$)
$\frac{1}{x}={\log }_{1000}2.3$ ......(i)
Similar way Consider $(0.23{)}^y=1000$
${\log }_{0.23}1000=y$
$\frac{1}{{\log }_{1000}0.23}=y$ (Since, ${\log }_{m}n=\frac{1}{{\log }_{n}m}$)
$\frac{1}{y}={\log }_{1000}0.23$.....(ii)
Consider $\frac{1}{x}-\frac{1}{y}={\log }_{1000}2.3-{\log }_{1000}0.23$ (from equation (i) and (ii))
$={\log }_{{10}^3}\frac{2.3}{0.23}$
$=\frac{1}{3}{\log }_{10}10$ (Since, ${\log }_{a^m}b=\frac{1}{m}{\log }_{a}b$)
$\therefore \frac{1}{x}-\frac{1}{y}=\frac{1}{3}$ (Since, ${\log }_{a}a=1$)