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

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henkjan Group TitleBest ResponseYou've already chosen the best response.1
log(x+2)+log(x1) = log ((x+2)(x1)) maybe that helps ;P
 2 years ago

henkjan Group TitleBest ResponseYou've already chosen the best response.1
And 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.
 2 years ago

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

Confucious Group TitleBest ResponseYou've already chosen the best response.0
but it's equal to one??, in the first line you have the equation equal to 10
 2 years ago

henkjan Group TitleBest ResponseYou've already chosen the best response.1
I 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
 2 years ago

Confucious Group TitleBest ResponseYou've already chosen the best response.0
isn't the b 10?
 2 years ago

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

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

henkjan Group TitleBest ResponseYou've already chosen the best response.1
log(x)+log(y) = log(x*y) if log(a)=b then 10^b = a only 2 rules I used
 2 years ago

henkjan Group TitleBest ResponseYou've already chosen the best response.1
yes, indeed than it would be a = 10^2 = 100
 2 years ago

Confucious Group TitleBest 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?
 2 years ago

henkjan Group TitleBest ResponseYou've already chosen the best response.1
oh, 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
 2 years ago

Confucious Group TitleBest ResponseYou've already chosen the best response.0
oh okay, thank you very much
 2 years ago
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