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ln(x)+ln(x+2)=ln(8) i started with ln(x(x+2)=ln(8) so I have ln(x^2+2x)=ln(8) what next?
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subtract ln 8 from both sides ln (x^2 + 2x) - ln 8 = 0
do you see the next step now?
I tryed that then it would be ln (x^2+2x)/8 then how do i solve from there
ln (x^2 +2x)/8 = 0 that = 0 is very important
now you turn it into exponential form...ln has a base e right? so in exponential form it becomes \[e^0 = \frac{x^2+2x}{8}\] agree?
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ok i got that so e^0 equals 1 so then i can just solve for x?
yup
and it cannot be negative correct?
yep..because you have an ln x
thanks
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x+2 cannot be negative or equal to 0 as well so x has to be greater than 2
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