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log(x+2)+log(x-1) = log ((x+2)(x-1)) maybe that helps ;P
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.
So (x+2)(x-1) = 10^1 x^2+x-2= 10 x^2+x-12 = 0 Solve for x and find x = -4 and x = 3
but it's equal to one??, in the first line you have the equation equal to 10
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 = (x-1)(x+2) in this case
isn't the b 10?
no, the b is equal to the right side in this equation log ((x+2)(x-1)) =1 ...... so it's 1
so if the equation was equal to 2 the b would be 10^2?
log(x)+log(y) = log(x*y) if log(a)=b then 10^b = a only 2 rules I used
yes, indeed than it would be a = 10^2 = 100
:) alright, the answer at the back of the text is just 3, why isn't the -4 acceptable?
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
oh okay, thank you very much