Question

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

Answer 1

given (2.3)x=1000
3 3
(23÷10)x=10 (10 = 1000)
3
(23)x÷10x=10
3
(23)x=10 ×10x
3+x
(23)x=10 (since if bases are equal then powers also equal)
3+x÷x
Let 23=10 as equation (1)

(0.23)y=1000
3
(23÷100)y = 10
2 3
(23)y÷(10 )y=10
3+2y
(23)y=10 × 10
3+2y÷y
Let (23)y=10 as equation (2)

from equation (1)&(2)
3+x÷x. 3+2y÷y
10 = 10 [if bases are equal then powers also equal]

3+x÷x=3+2y÷y

3÷x+x÷x=3÷y+2y÷y

3÷x+1=3÷y+2

3÷x-3÷y=2-1

3[1÷x-1÷y]=1

Therefore 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