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

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

henkjan
 3 years ago
Best ResponseYou've already chosen the best response.1And 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.

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

Confucious
 3 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

henkjan
 3 years ago
Best ResponseYou've already chosen the best response.1I 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

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

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

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

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

Confucious
 3 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?

henkjan
 3 years ago
Best ResponseYou've already chosen the best response.1oh, 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

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