SOLUTION: Log2x+log2(x+2)=log2(x+6)

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Question 256861: Log2x+log2(x+2)=log2(x+6)
Found 2 solutions by drk, CharlesG2:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Here is the question:
Log_2%28x%29+%2B+log_2%28x%2B2%29+=+log_2%28x%2B6%29
since all are base 2 and we see + on the left, we can rewrite as
log_2%28x%29%28x%2B2%29+=+log_2%28x%2B6%29
now cross out the log_2 and solve for x as
x%28x%2B2%29+=+x%2B6
x%5E2+%2B+2x+-+x+-+6+=+0
x%5E2+%2B+x+-+6+=+0
%28x%2B3%29%28x-2%29=0
So, x = -3 or x = 2.
If x = -3, then there would be no solution.
So, x = 2.

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
Log2x+log2(x+2)=log2(x+6)
log(2x)+log(2*(x+2))=log(2*(x+6))
logb (mn) --> b=base --> logb (mn) = logb (m) + logb (n)
in this problem's case the base b is 10
log(2x)+log(2*(x+2))=log(2*(x+6))
log(2x*2*(x+2))=log(2*(x+6))
log(4x*(x+2))=log(2*(x+6))
log(4x^2+8x)=log(2x+12)
4x^2+8x=2x+12
4x^2+6x-12=0
b^2-4*a*c=6^2-4*4*(-12) --> a=4,b=6,c=-12
36-16*(-12)
36-(-192)
36+192
228
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-6+%2B-+sqrt%28+228+%29%29%2F8+
x+=+%28-6+%2B-+sqrt%28+228+%29%29%2F8+
228=2*114=2*2*57
x+=+%28-6+%2B-+2%2Asqrt%2857%29%29%2F8+
x+=+%28-3+%2B-+sqrt%2857%29%29%2F4+ (took out 2/2)
x+=+%28-3%2F4+%2B-+sqrt%2857%29%2F4%29+
x+=+%28-0.75+%2B-+1.887459%29+ (rounded the root/4 to 6 places)
x1=1.137459
x2=-2.637459
but going back to Log2x+log2(x+2)=log2(x+6)
we can not use x2 since we can not take the log of a negative number
since logb y = x is same as b^x=y, you would be asking what to the what is a negative number
so x is x1 or 1.137459