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May 5th 2006, 12:48 AM
kenj

Logs

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

Quote:

Originally Posted by kenj

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:

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

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