Question

If (2.3)x =(0.23)y =1000 then find the value of 1_x -1_y
​

Answer 1

Answer:

1/x - 1/y = 1/3

Explanation:

(2.3)x = (0.23)y = 1000 

=> log10(2.3)^y = log10(0.23)^y = log10(10³)

=> x log10(2.3) = ylog10(0.23) = 3

=> 3/x = log10 2.3, 3/y = log10 0.23

=> 3(1/x - 1/y) = 1

=> 1/x - 1/y = 1/3

Answer 2

Answer:

1/3

Step-by-step explanation:

(2.3)^x=1000 ------>(1)

(0.23)^y=1000 ----->(2)

(2.3)^x=1000

Apply log on both sides

log (2.3)^x=log1000

x log(2.3)=1000

1/x=log(2.3)/log1000

(0.23)^y=1000

Apply log on both sides

log(0.23)^y=log1000

y log(0. 23)=log1000

1/y=log(0.23)/log1000

1/x-1/y=log(2.3)/log1000-log(0.23)/log1000

log(2.3)-log(0.23)/log1000

log(2.3)-log2.3/10/1000

log(2. 3)-log2.3-log10/log1000

log (2.3) get cancelled and we get log 10/log1000

log10/log10^3

=log 10 get cancelled we get 1/3 .