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

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

Mathematics
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|dw:1437291202065:dw|
we need our three log rules \[\log_b(mn) =\log_bm+\log_bn\] \[\log_b(\frac{m}{n}) =\log_bm-\log_bn\] \[\log_b(m^n) =n \cdot \log_bm\]
so for this problem we don't need the subtraction one.

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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
|dw:1437293044978:dw| so use the third rule first on this
use \[\log_b(m^n) =n \cdot \log_bm \] which is the log rule for exponents first on your drawing
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?
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.
OMG apologizes typing too fast \[Log_b(\frac{XY}{Z})\]
Logb(3(2)/b)?
but you just have addition for this though so the second rule isn't necessary |dw:1437293682407:dw|
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
|dw:1437293771745:dw|
\[\log_b(m^n) =n \cdot \log_bm \] use the left side of this rule m = x n = 3 for the first m = y n = 2 for the second
I'll do one and you can do the other one using exponent rule for logs |dw:1437294023458:dw|
|dw:1437294470191:dw|
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
|dw:1437294629036:dw|
whoa we do not combine exponents... but the xy part is correct.
|dw:1437294775553:dw|
the exponents couldn't be combined anyway due to different variables
so its just logb(x^3y^2)?
yes |dw:1437294894273:dw|
thank you

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