solve for first x
10^y=(x-2)
x= 10^y +2
solve for the second x
y-1=log(x+1)
10^(y-1)=x+1 or x= 10^(y-1) +1
the x=x, y=y
10^(y-1)+1=10^y +2
remember a^(b-c)= a^b/a^c so
10^y/10 + 1=10^y +2
10^y (-9/10)=1
or -9*10^y=10
log of each side
-9 +y=1
y=10
so, now figure x..
x= 10^(y-1) +1=10^9+1
weird. Check my work.
determine the point of intersection
y=logbase 10(x-2)
and y=1-logbase 10(x+1)
3 answers
you are THE BESTT!!
its wrong
you wrote
10^(y-1)+1=10^y +2
its supposed to be
10^(y-1)-1=10^y +2
THE ANSWER IS APPARENTLY 2, log base 10 2
you wrote
10^(y-1)+1=10^y +2
its supposed to be
10^(y-1)-1=10^y +2
THE ANSWER IS APPARENTLY 2, log base 10 2