Question

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

Answer 1

easy one ,
Given ,
(2.3)^x = (0.23)^y = 1000
First ,
(2.3)^x = 1000
taking log on bothsides,
log (2.3)^x = log1000
x log (2.3) = 3
x ×0.361727 = 3
x = 8.2935

Second,
(0.23)^y = 1000
taking log on bothsides

log (0.23)^y = log (1000)
y log (0.23) = 3
y×(-0.63827) = 3
y = -4.7

Now, 1/x-1/y = (1/8.2935)-(1/-4.7)
= 1/8.3+1/4.7
= (4.7+8.3)/(8.3×4.7)
= 13/39
= 0.33333 is right answer.
hope it helps you

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