Respuesta :

konrad509

[tex]\log x + \log y = \log\dfrac{1}{x}\\\\\log y = \log\dfrac{1}{x}-\log x\\\\\log y = \log\dfrac{1}{x^2}\\\\y=\dfrac{1}{x^2}[/tex]

Answer:

y = 1/x^2.

Step-by-step explanation:

log x + log y = log(1/x)

log y = log(1/x) - log x

Using the rule  log a - log b =  log a/b, we have:

log y = log ( 1/x  / x)

log y = log (1/x^2)

So y = 1/x^2.