anonymous
  • anonymous
expanding logs
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
condense \[1/3 logx-2\log(x+2)-4\log(y+2)\]
phi
  • phi
move the coefficients inside the logs use the rule \[ a \log(b) = \log(b^a) \]
anonymous
  • anonymous
does it matter if in base of three

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phi
  • phi
no.
anonymous
  • anonymous
so log1/3x-2log2x+4-log4y+8
phi
  • phi
try again the coefficient becomes an *exponent* example: the first log becomes \[ \log(x^\frac{1}{3}) \]
anonymous
  • anonymous
\[logx ^{1/3}-Log(x+2)^{2}-Log(y+2)^{4}\] \ \
phi
  • phi
ok, now if all the logs are the same base you can use the rule \[ \log(a) - \log(b)\ = \log(\frac{a}{b} )\]
anonymous
  • anonymous
but can i have loga-logb-logc= log a/b/c
anonymous
  • anonymous
the three logs are throwing me off
phi
  • phi
yes. a/b/c is the same as a/(b*c) another way to do this is factor out the minus sign \[ \log(x^(1/3) - (\log( (x+2)^2 ) +\log( (y+2)^4) )\] and combine the logs inside the parens using log(a)+log(b)= log(a*b)
anonymous
  • anonymous
okso...|dw:1378675998926:dw|
anonymous
  • anonymous
terrible drawing
phi
  • phi
yes, but no -1 out front. when you do log(a) - (log(hairy) ) = log(a/hairy) no - out front
anonymous
  • anonymous
so no -1 ok
anonymous
  • anonymous
yhx
anonymous
  • anonymous
t
anonymous
  • anonymous
1 more i will post
phi
  • phi
oh wait, one mistake. in the denominator, you *multiply* (not add)

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