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Math Help - Log Simultaneous Equation Help

  1. #1
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    Log Simultaneous Equation Help

    I'm trying to determine the equation of a graph, and I have come up with these two equations, how do I solve for a and b?

    2=a\log_2(5-b)
    4=a\log_2(7-b)
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  2. #2
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    Quote Originally Posted by user_5 View Post
    I'm trying to determine the equation of a graph, and I have come up with these two equations, how do I solve for a and b?

    2=a\log_2(5-b)
    4=a\log_2(7-b)
    The simplest way to eliminate a from those equations is to divide one by the other:
    \frac{4}{2}= 2= \frac{log_2(7- b)}{log_2(5- b)}
    or simply
    2\log_2(5- b)= log_2(7- b)
    log_2((5-b)^2= log_2(7- b)

    and, since logarithm is "one-to-one", (5- b)^2= 7- b, a quadratic equation for b.
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  3. #3
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    Hello, user_5!

    Another approach . . .


    I'm trying to determine the equation of a graph. .**
    I have come up with these two equations.
    How do I solve for a and b?

    . . \begin{array}{cccc}2&=& a\log_2(5-b)  & [1] \\ \\[-3mm] 4 &=& a\log_2(7-b) & [2] \end{array}

    \begin{array}{cccccc}<br />
\text{Multiply [1] by 2:} & 2a\log_2(5-b) &=& 4  \\<br />
\text{Equate to [2]:} & a\log_2(7-b) &=& 4 \end{array}


    We have: . 2{\color{red}\rlap{/}}a\log_2(5-b) \;=\;{\color{red}\rlap{/}}a\log_2(7-b) \quad\Rightarrow\quad \log_2(5-b)^2 \:=\:\log_2(7-b)

    Then: . (5-b)^2 \:=\:7-b \quad\Rightarrow\quad b^2 -9b + 18\:=\:0

    . . . . (b-3)(b-6) \:=\:0 \quad\Rightarrow\quad b \:=\:3,\,6


    But b=6 is extraneous.

    The only solution is: . \boxed{b = 3}


    Substitute into [1]:

    . . a\log_2(5-3)\:=\:2 \quad\Rightarrow\quad a\log_2(2) \:=\:2 \quad\Rightarrow\quad a\cdot 1 \:=\:2 \quad\Rightarrow\quad\boxed{ a \:=\:2}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    You are given the function: . f(x) \;=\;\log_2(x-b)

    . . and two points: . (5,2),\;(7,4)

    Right?

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