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Thread: Logs

  1. #1
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    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))
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  2. #2
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    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:

    $\displaystyle y=a^x+1,\ a>2$, 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
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  3. #3
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    Sorry - got it ! It was easy once I realised that a^(x+1) is the same as aa^x
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