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Annetta_Martin
 one year ago
Help with Logarithms?
Annetta_Martin
 one year ago
Help with Logarithms?

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nathan917
 one year ago
Best ResponseYou've already chosen the best response.0Is there anymore to the question?

Annetta_Martin
 one year ago
Best ResponseYou've already chosen the best response.0These are some problems I did before it.....

Annetta_Martin
 one year ago
Best ResponseYou've already chosen the best response.01. 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

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.2dw:1365379791371:dw

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.2dw:1365379863911:dw

electrokid
 one year ago
Best ResponseYou've already chosen the best response.0they are inverse of eachother
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