Nick88888888
  • Nick88888888
Write as one logarithm with base b: 3 logb x + 2 logb y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Nick88888888
  • Nick88888888
|dw:1437291202065:dw|
UsukiDoll
  • UsukiDoll
we need our three log rules \[\log_b(mn) =\log_bm+\log_bn\] \[\log_b(\frac{m}{n}) =\log_bm-\log_bn\] \[\log_b(m^n) =n \cdot \log_bm\]
UsukiDoll
  • UsukiDoll
so for this problem we don't need the subtraction one.

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UsukiDoll
  • UsukiDoll
so anything to do with addition as seen in your problem will become multiplication as the end result... but first we need to use the third rule because you got exponents
UsukiDoll
  • UsukiDoll
|dw:1437293044978:dw| so use the third rule first on this
UsukiDoll
  • UsukiDoll
use \[\log_b(m^n) =n \cdot \log_bm \] which is the log rule for exponents first on your drawing
Nick88888888
  • Nick88888888
would using Log x + Log y - Log z = Solution: Log (xy)/z work? Yea i know it doesnt have z but would this method work?
UsukiDoll
  • UsukiDoll
ummm yeah but where is the base though? like ok... let's have the base be b \[Log_b X + Log_b Y - Log_bZ \] then it's a combination of the first and second log rule. Your final answer should look like \[Log_b(\frac{xy}{b})\] because we have a + sign which means multiplication according to the log rules and we hae a - sign which means division according to the log rules as well.
UsukiDoll
  • UsukiDoll
OMG apologizes typing too fast \[Log_b(\frac{XY}{Z})\]
Nick88888888
  • Nick88888888
Logb(3(2)/b)?
UsukiDoll
  • UsukiDoll
but you just have addition for this though so the second rule isn't necessary |dw:1437293682407:dw|
UsukiDoll
  • UsukiDoll
on top of that \[\log_b(m^n) =n \cdot \log_bm \] <-- this exponent rule comes first \[\log_b(mn) =\log_bm+\log_bn \] <-- this addition rule comes second
UsukiDoll
  • UsukiDoll
|dw:1437293771745:dw|
UsukiDoll
  • UsukiDoll
\[\log_b(m^n) =n \cdot \log_bm \] use the left side of this rule m = x n = 3 for the first m = y n = 2 for the second
UsukiDoll
  • UsukiDoll
I'll do one and you can do the other one using exponent rule for logs |dw:1437294023458:dw|
Nick88888888
  • Nick88888888
|dw:1437294470191:dw|
UsukiDoll
  • UsukiDoll
alright.... now we just need the first exponent rule... so since we have addition our final result should be multiplication |dw:1437294541977:dw| fill in the blank
Nick88888888
  • Nick88888888
|dw:1437294629036:dw|
UsukiDoll
  • UsukiDoll
whoa we do not combine exponents... but the xy part is correct.
UsukiDoll
  • UsukiDoll
|dw:1437294775553:dw|
UsukiDoll
  • UsukiDoll
the exponents couldn't be combined anyway due to different variables
Nick88888888
  • Nick88888888
so its just logb(x^3y^2)?
UsukiDoll
  • UsukiDoll
yes |dw:1437294894273:dw|
Nick88888888
  • Nick88888888
thank you

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