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))

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- May 5th 2006, 12:48 AMkenjLogs
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 AMCaptainBlackQuote:

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