# Logs

• May 5th 2006, 12:48 AM
kenj
Logs

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))
• May 5th 2006, 02:53 AM
CaptainBlack
Quote:

Originally Posted by kenj

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:

$y=a^x+1,\ a>2$, and
$y=a^{x+1},\ a>2$

intersect when:

$
a^x+1=a^{x+1}
$

so:

$
a^x+1=a.a^x
$

Rearranging:

$
a^x(a-1)=1
$
,

rearranging:

$
a^x=\frac{1}{a-1}
$

so taking $\log$s to base $a$:

$
x=\log_a(1/(a-1))=-\log_a(a-1)
$

RonL
• May 5th 2006, 03:09 AM
kenj
Sorry - got it ! It was easy once I realised that a^(x+1) is the same as aa^x