A point of intersection is when the two equations equal each other. So set:
log(x-2) = 1 - log(x+1)
Then solve for x.
Does that help, or are the logarithms themselves giving you trouble?
determine the point of intersection
y=logbase 10(x-2)
and y=1-logbase 10(x+1)
2 answers
log(x-2)= 1-log(x+1)
log(x-2)+log(x+1)=1
log(a)+log(b)=log(a*b)
log(x-2)+log(x+1)=log((x-2)*(x-1))
1=log(10)
log((x-2)*(x-1))=log(10)
So:
(x-2)*(x+1)=10
x^2-2x+x-2=10
x^2-x-2=10
x^2-x-2-10=0
x^2-x-12=0
Exact solutions of this equation is:
-3 and 4
Negative numbers have not logarithms.
So solution is x=4
log(x-2)+log(x+1)=1
log(a)+log(b)=log(a*b)
log(x-2)+log(x+1)=log((x-2)*(x-1))
1=log(10)
log((x-2)*(x-1))=log(10)
So:
(x-2)*(x+1)=10
x^2-2x+x-2=10
x^2-x-2=10
x^2-x-2-10=0
x^2-x-12=0
Exact solutions of this equation is:
-3 and 4
Negative numbers have not logarithms.
So solution is x=4