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

Question 256861: Log2x+log2(x+2)=log2(x+6)
Found 2 solutions by drk, CharlesG2:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Here is the question:

since all are base 2 and we see + on the left, we can rewrite as

now cross out the log_2 and solve for x as

So, x = -3 or x = 2.
If x = -3, then there would be no solution.
So, x = 2.
Answer by CharlesG2(834) (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

228=2*114=2*2*57

(took out 2/2)

(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

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