.Sam.
  • .Sam.
1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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.Sam.
  • .Sam.
\[\huge \log_{2}x-\log_{4}y=4\] \[\huge \log_{2}(x-2y)=5\]
anonymous
  • anonymous
log(2) x - log(4) y = 4 log(2) x - [log(2) y]/[log(2) 4] = 4 log(2) x - [log(2) y]/2 = 4 [2 log(2) x]/2 - [log(2) y]/2 = 4 log(2) x^2 - log(2) y = 8 log(2) x^2/y = 8 x^2/y = 2^8________________(1) log(2) x-2y = 5 x - 2y = 2^5 x = 2y + 32________________(2) substitute equation (2) into equation (1); (2y + 32)^2/y = 2^8 (2y + 32)^2 = 256y 4y^2 + 128y + 1024 = 256y 4y^2 -128y + 1024 = 0 y^2 - 32y + 256 = 0 (y - 16)(y - 16) = 0 y = 16 substitute y = 16 into equation (2); x = 2(16) + 32 x = 64
lgbasallote
  • lgbasallote
2^5 = x - 2y 32 = x -2y 32 + 2y = x..that's a good start

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anonymous
  • anonymous
@lgbasallote give medal
lgbasallote
  • lgbasallote
log_2 (32 + 2y) - log_4 (y) = 4 ahhh if only i remember how to turn log_4 to log_2 or vice versa -__-
lgbasallote
  • lgbasallote
hmmm depends...is rohan right @.sam.?
anonymous
  • anonymous
very dumb and easiest problem
.Sam.
  • .Sam.
@Rohangrr don't use offensive languages, and I don't understand what you're working
anonymous
  • anonymous
@.Sam. I'm really soooooooory well yes its the easiest one
lgbasallote
  • lgbasallote
well..the hint says it's a hard one so i guess it cant be solved simply..there must be something else here..hmmm.
anonymous
  • anonymous
@Igbasallote i agree to u

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