Question

if (2.3)x=(0.23)y=1000 then find the value of 1/x-1/y.

Answer 1

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.

Answer 2

(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

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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