If x-log y=x+log y, where x and log y are real, then:

1972 AMC 12 Problem 8

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\lvert x-\log y\rvert=x+\log y,

where

x

and

\log y

are real, then:

x=0

y=1

x=0

and

y=1

x(y-1)=0

None of these

Separate the cases

x-\log y\ge0

and

x-\log y\lt0

One case forces

\log y=0

, while the other forces

x=0

x-\log y\ge0,

the equation gives

\log y=0,

hence

y=1.

x-\log y\lt0,

it gives

x=0.

Therefore every solution satisfies either

x=0

y=1,

which is expressed by

x(y-1)=0.

Therefore, the correct answer is D.