Question

if 2.3^x=0.23^y=1000 then 1/x-1/y value

Answer 1

$2.3^x=0.23^y=1000\\ \\ Taking\ log\ \\ \\ log(2.3^x)=log(0.23^y)=log(1000)\\ \\ \Rightarrow xlog(2.3)=ylog(0.23)=log(10^3)\\ \\ \Rightarrow xlog(0.23 \times 10)=ylog(0.23)=3log(10)\ \\\\ \Rightarrow x[log(0.23)+log (10)]=ylog(0.23)=3log(10)\ \\\\ \Rightarrow x[log(0.23)+1]=ylog(0.23)=3$

$\\\\ \Rightarrow x= \frac{3}{ log(0.23)+1}\ \ \ and\ \ \ \ y= \frac{3}{log(0.23)}$


$\frac{1}{x}- \frac{1}{y}= \frac{ log(0.23)+1}{3}-\frac{log(0.23)}{3}=\frac{ log(0.23)+1-log(0.23)}{3}=\boxed{ \frac{1}{3} }$

Answer 2

Answer :-

$\\ \\ \tt \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{3}$

Solution :-

$\sf {(2.3)}^{x} = {(0.23)}^{y} = 1000$

$\\ \sf (i) \: {(2.3)}^{x} = 1000 \\ \\ \\ \sf \longrightarrow {(2.3)}^{x} = {10}^{3} \\ \\ \\ \sf \longrightarrow 2.3 = \sqrt[x]{ {10}^{3} } \\ \\ \\ \large \sf \longrightarrow 2.3 = {10}^{ \frac{3}{x} } - - - (1)$

$\\ \\ \sf (ii) {(0.23)}^{y} =1000 \\ \\ \\ \sf \longrightarrow {(0.23)}^{y} = {10}^{3} \\ \\ \\ \sf \longrightarrow 0.23 = \sqrt[y]{ {10}^{3} } \\ \\ \\ \large \sf \longrightarrow 0.23 = {10}^{ \frac{3}{y} }$

Multiplying by 10 on both sides

$\\ \\ \longrightarrow \large \sf 0.23 \times 10 = {10}^{ \frac{3}{y} } \times 10 \\ \\ \\ \large \sf \longrightarrow 2.3 = {10}^{ \frac{3}{y} + 1 } - - (2) \\ \\ \\ \boxed{ \bf \because {a}^{m} \times {a}^{n} = {a}^{m + n} }$

From eq.(1) an eq.(2)

$\\ \\ \large \sf \longrightarrow {10}^{ \frac{3}{x} } = {10}^{ \frac{3}{y} + 1} \\ \\ \\ \sf \longrightarrow \dfrac{3}{x} = \dfrac{3}{y} + 1 \\ \\ \\ \boxed{ \bf \because {a}^{m} = {a}^{n} \: then \: m = n} \\ \\ \\ \sf \longrightarrow \dfrac{3}{x} - \dfrac{3}{y} = 1 \\ \\ \\ \sf \longrightarrow 3 \left( \dfrac{1}{x} - \dfrac{1}{y} \right) = 1 \\ \\ \\ \bf \longrightarrow \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{3}$