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Nick88888888
 one year ago
Write as one logarithm with base b: 3 logb x + 2 logb y
Nick88888888
 one year ago
Write as one logarithm with base b: 3 logb x + 2 logb y

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Nick88888888
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437291202065:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2so for this problem we don't need the subtraction one.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437293044978:dw so use the third rule first on this

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2use \[\log_b(m^n) =n \cdot \log_bm \] which is the log rule for exponents first on your drawing

Nick88888888
 one year ago
Best ResponseYou've already chosen the best response.0would 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
 one year ago
Best ResponseYou've already chosen the best response.2ummm 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
 one year ago
Best ResponseYou've already chosen the best response.2OMG apologizes typing too fast \[Log_b(\frac{XY}{Z})\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but you just have addition for this though so the second rule isn't necessary dw:1437293682407:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2on 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437293771745:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\log_b(m^n) =n \cdot \log_bm \] use the left side of this rule m = x n = 3 for the first m = y n = 2 for the second

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I'll do one and you can do the other one using exponent rule for logs dw:1437294023458:dw

Nick88888888
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437294470191:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2alright.... 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437294629036:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2whoa we do not combine exponents... but the xy part is correct.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437294775553:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the exponents couldn't be combined anyway due to different variables

Nick88888888
 one year ago
Best ResponseYou've already chosen the best response.0so its just logb(x^3y^2)?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yes dw:1437294894273:dw
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