Question 256861
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-6 +- sqrt( 228 ))/8 }}}
{{{x = (-6 +- sqrt( 228 ))/8 }}}
228=2*114=2*2*57
{{{x = (-6 +- 2*sqrt(57))/8 }}}
{{{x = (-3 +- sqrt(57))/4 }}} (took out 2/2)
{{{x = (-3/4 +- sqrt(57)/4) }}}
{{{x = (-0.75 +- 1.887459) }}}  (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