Any help please?
Line 1: y = a^x + 1 , a>2
Line 2: y = a^(x+1), a>2
Prove that they intersect when x = loga(1/(a-1))
The curves:Originally Posted by kenj
$\displaystyle y=a^x+1,\ a>2$, and
$\displaystyle y=a^{x+1},\ a>2$
intersect when:
$\displaystyle
a^x+1=a^{x+1}
$
so:
$\displaystyle
a^x+1=a.a^x
$
Rearranging:
$\displaystyle
a^x(a-1)=1
$,
rearranging:
$\displaystyle
a^x=\frac{1}{a-1}
$
so taking $\displaystyle \log$s to base $\displaystyle a$:
$\displaystyle
x=\log_a(1/(a-1))=-\log_a(a-1)
$
RonL