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anonymous
 5 years ago
I'm not good with logs, can someone help me?
log(x+2)+log(x1)=1
anonymous
 5 years ago
I'm not good with logs, can someone help me? log(x+2)+log(x1)=1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log(x+2)+log(x1) = log ((x+2)(x1)) maybe that helps ;P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And perhaps another helpful rule: if log(a)=b than 10^b = a only in the case when 10 is the basenumber of the logaritm, but if it is not stated otherwise it is always 10.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So (x+2)(x1) = 10^1 x^2+x2= 10 x^2+x12 = 0 Solve for x and find x = 4 and x = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it's equal to one??, in the first line you have the equation equal to 10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I applied the rule I explained in my second post here. because b is in this case 1, it means that 10^1 = a a = (x1)(x+2) in this case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, the b is equal to the right side in this equation log ((x+2)(x1)) =1 ...... so it's 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if the equation was equal to 2 the b would be 10^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log(x)+log(y) = log(x*y) if log(a)=b then 10^b = a only 2 rules I used

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, indeed than it would be a = 10^2 = 100

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0:) alright, the answer at the back of the text is just 3, why isn't the 4 acceptable?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, right.... a logaritmic can never have a negative number log(x) is not possible... as 10^b = a ... a can only become positive in this formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay, thank you very much
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