anonymous
  • anonymous
Help with Logarithms?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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nathan917
  • nathan917
Is there anymore to the question?
anonymous
  • anonymous
These are some problems I did before it.....

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anonymous
  • anonymous
1. log7 + log (n - 2) = log 6n log [7(n - 2)] = log (6n) 7(n - 2) = 6n n = 14 2. log5 (16) - log5 (2t) = log5 (2) log5 (16 / 2t) = log5 (2) 16 / 2t = 2 t = 4 3. log5 m = (1/3) log5 (125) log5 (m) = log5 [(125^(1/3)] m = 125^(1/3) m = 5 4. log y = (1/4) log (16) + (1/2) log (49) log y = log [16^(1/4)] + log [49^(1/2)] =log y = log (2) + log (7) log y = log (2 . 7) y = 14 5. log6 (b^2 + 2) + log6 (2) = 2 log6 [(b^2 + 2) . 2] = 2 (b^2 + 2) . 2 = 6² b^2 + 2 = 18 b^2 = 16 b = 4 or b = - 4 6. log3 (5x + 5) - log3 (x^2 - 1) = 0 log3 [(5x + 5) / (x^2 - 1)] = 0 (5x + 5) / (x^2 - 1) = 3^0 5x + 5 = x² - 1 x² - 5x - 6 = 0 x = 6 or x = - 1 x = 6 7. log2 (x - 2) + 5 = 8 - log2 (4) log2 (x - 2) + 5 = 8 - 2 log2 (x - 2) = 1 x - 2 = 2^1 x = 4
anonymous
  • anonymous
my eyes !!!!
anonymous
  • anonymous
|dw:1365379791371:dw|
anonymous
  • anonymous
|dw:1365379863911:dw|
anonymous
  • anonymous
they are inverse of eachother

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