I have a problem, solve for x.
ln(x+2)-2ln(3)=0 ln(x+2)-2ln3=0
So I guess 2ln(3)=2ln
But first I did this because I didn't notice the second equation.
ln(x+2)=2ln(3)
[e^(2ln(3)]=x+2
x=[e^2ln(3)]-2
What do you think?
I have a problem, solve for x.
ln(x+2)-2ln(3)=0 ln(x+2)-2ln3=0
So I guess 2ln(3)=2ln
But first I did this because I didn't notice the second equation.
ln(x+2)=2ln(3)
[e^(2ln(3)]=x+2
x=[e^2ln(3)]-2
What do you think?