Jeanettaanglero Jeanettaanglero
  • 11-06-2021
  • Mathematics
contestada

I need help please!!!

I need help please class=

Respuesta :

Аноним Аноним
  • 11-06-2021

Answer:

1

Step-by-step explanation:

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wegnerkolmp2741o
wegnerkolmp2741o wegnerkolmp2741o
  • 11-06-2021

Answer:

log x^2 sqrt(y)

Step-by-step explanation:

1/2 log y + 2 logx

We know that a log b = log b^2

log y^ 1/2 + log x^2

We know that log a + log b = log (ab)

log  x^2 y^ 1/2

log x^2 sqrt(y)

Answer Link

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