Korey1998
Korey1998 Korey1998
  • 12-07-2016
  • Mathematics
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I have no clue how to do this...

I have no clue how to do this class=

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dalendrk
dalendrk dalendrk
  • 13-07-2016
[tex]\log16x^2+2\log\dfrac{1}{x}=\log(4x)^2+2\log\dfrac{1}{x}=2\log4x+2\log\dfrac{1}{x}\\\\=2\left(\log4x+\log\dfrac{1}{x}\right)=2\log\left(4x\cdot\dfrac{1}{x}\right)=2\log4=\log4^2=\boxed{\log16}[/tex]
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