Question

If log₂(x + y) = log₃(x - y) = `log 25/log 0.2`, find the values of x and y.

Answer 1

log₂(x + y) = `log 25/log 0.2` 

⇒ log₂(x + y) = log_0.2 25

⇒ log₂(x + y) = `log_(2/10) 25`

⇒ `log_2( x + y ) = log_5^-1 5^2`

⇒ `log_2( x + y ) = -2log_5 5`

⇒ `log_2( x + y ) = -2`

⇒ x + y = 2^-2                 ...[Removing logarithm]

⇒ x + y = `1/2^2`

⇒ x + y = `1/4`               ...(1)

⇒ `log_3( x - y ) = log25/log 0.2`

⇒ `log_3( x - y ) = log5^2/log5^1`

⇒ `log_3(x - y) = (2log5)/(-1log5)`

⇒ `log_3( x - y ) = -2`          

3^-2 = x - y

`1/3^2=x-y`

`1/3=x-y`

x - y = `1/9`          ...(2) 

Adding equation (1) and (2)

x + y = `1/4`

x - y = `1/9`

2x = `1/4+1/9`

2x = `(9+4)/36`

2x = `13/36`

x = `13/72`

From equation (1)

`13/72+y=1/4`

y = `1/4-13/72`

y = `(18-13)/72`

y = `5/72`