Plz answer this correctly
Answers
Step-by-step explanation:
Given that,
2.3^x=0.23^y=1000
Now,
2.3^x=1000 and 0.23^y=1000
Applying log on both sides we get,
log2.3^x=log1000
xlog2.3=log10^3
xlog(23/10)=3
log(23/10)=3/x.....................................1
and,
log0.23^y=log1000
ylog(23/100)=log10^3
ylog(23/100)=3
log(23/100)=3/y..................................2
By eq1-eq2 we get,
log(23/10)-log(23/100)=3/x-3/y
log{(23/10)/(23/100}=3(1/x-1/y)
log10=3(1/x-1/y)
1=3(1/x-1/y)
1/3=1/x-1/y.
Hence value of 1/x-1/y=1/3.
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(2.3)^x=1000 and (0.23)^y=1000
log(2.3)^x =log1000 and log(0.23)^y=1000
xlog(2.3)=log10^3 and log(0.23)^y =log10^3
xlog(2.3) = 3log10 and ylog(0.23) =3log10
xlog(2.3) = 3×1 and ylog(0.23)=3 [log10=1]
⇒log(2.3)=3/x and log0.23 =3/y
/////////////
3/x -3/y =log2.3 -log(0.23)
3(1/x - 1/y) =log(2.3/0.23)
⇒1/x - 1/y ={log(10)}/3
⇒1/x-1/y =(1)/3
⇒1/x -1/y =1/3
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