Question

If (2.3)ˣ = (0.23)ⁿ= 1000 the find the value of 1/x-1/y

Answer 1

Given,
(2.3)^x=(0.23)^y=1000
We have,
(2.3)^x=1000
Applying log on both sides we get,
log(2.3)^x=log1000

xlog(23/10)=3

log(23/10)=3/x....................................1
and,
(0.23)^y=1000
Applying log on both sides we get,
log(0.23)^y=log1000

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/x-1/y=1/3

Hence value of 1/x-1/y=1/3.

Answer 2

Step-by-step explanation:

x(log.23+log10) = y log .23 =log 10³=3

1/x = log.23 +1 /3

1/y = log.23/3

let log.23 = a

1/x -1/y = (a+1)/3-(a/3)= (a+1-a)/3 =1/3