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Thread: Power Functions

  1. #1
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    Power Functions

    Find a power function through the two points: (7,8) (1,0.7)
    $\displaystyle f(x) = kx^p$

    This is what I've gotten only:
    $\displaystyle 8 = k7^p$
    $\displaystyle 0.7 = k1^p$

    I tried solving for p but it came out to be -1.25 and not 1.25 like the back of the book.

    Thank you!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by hydride View Post
    Find a power function through the two points: (7,8) (1,0.7)
    $\displaystyle f(x) = kx^p$

    This is what I've gotten only:
    $\displaystyle 8 = k7^p$
    $\displaystyle 0.7 = k1^p$

    I tried solving for p but it came out to be -1.25 and not 1.25 like the back of the book.

    Thank you!
    take logs so $\displaystyle y=kx^p$ becomes:

    $\displaystyle \log(y)=p \log(x)+k$

    Now the problem becomes that of finding the line through the points $\displaystyle (\log(7),\log(8))$ and $\displaystyle (\log(1),\log(0.7))$

    CB
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  3. #3
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    Hello, hydride!

    Find a power function through the two points: (7, 8), (1, 0.7)

    $\displaystyle f(x) = kx^p$

    This is what I've gotten only:
    $\displaystyle 8 \:=\: k\!\cdot\!7^p$
    $\displaystyle 0.7 \:=\: k\!\cdot\!1^p$

    I tried solving for p but it came out to be -1.25 . . . . How?
    and not 1.25 like the back of the book.
    We have: .$\displaystyle \begin{array}{cccc}7^p\!\cdot\!k &=& 8 & [1] \\ k &=& 0.7 & [2] \end{array}$

    Divide [1] by [2]: .$\displaystyle \frac{7^p\!\cdot\!k}{k} \:=\:\frac{8}{0.7} \quad\Rightarrow\quad 7^p \:=\:\frac{80}{7} $

    Take logs, base 7: .$\displaystyle \log_7\left(7^p\right) \:=\:\log_7\!\left(\tfrac{80}{7}\right) \quad\Rightarrow\quad p\cdot\log_7(7) \:=\:\log_7\!\left(\tfrac{80}{7}\right) $

    . . Hence: .$\displaystyle p \:=\:\log_7(\tfrac{80}{7})$


    Substitute into [1]: .$\displaystyle 7^{\log_7(\frac{80}{7})}\!\cdot\!k \:=\:8 \quad\Rightarrow\quad \tfrac{80}{7}\,k \:=\:8 \quad\Rightarrow\quad k \,=\,0.7$


    Therefore: .$\displaystyle f(x) \;=\;0.7x^{\log_7(\frac{80}{7})} \;\approx\;0.7x^{1.25} $ .*


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

    *
    $\displaystyle \log_7\left(\frac{80}{7}\right) \;=\;\frac{\ln(\frac{80}{7})}{\ln7} \;=\;1.251916224 \;\approx\;1.25$

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  4. #4
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    How did you get k = .7?

    Can you do it an alternate way? By using substitution?
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  5. #5
    MHF Contributor

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    Here's how I would have done it:

    From $\displaystyle 8= k 7^p$ and $\displaystyle .7= k 1^p= k$ we have immediately that k= .7. Put that into the first equation: $\displaystyle 8= .7(7^p)$ so $\displaystyle 7^p= 8/.7= 11.42857$, approximately. Now take the logarithm of boht sides: $\displaystyle p log(7)= log(11.42857)$ and $\displaystyle p= \frac{log(11.42857)}{log(7)}$.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Here's how I would have done it:

    From $\displaystyle 8= k 7^p$ and $\displaystyle .7= k 1^p= k$ we have immediately that k= .7. Put that into the first equation: $\displaystyle 8= .7(7^p)$ so $\displaystyle 7^p= 8/.7= 11.42857$, approximately. Now take the logarithm of boht sides: $\displaystyle p log(7)= log(11.42857)$ and $\displaystyle p= \frac{log(11.42857)}{log(7)}$.
    what happend to the p when you figured out k = .7?
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  7. #7
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    when you figured out that k was .7 in the second set p is not needed becuase 1 ^x = 1
    you get the K and input it into the first set and get 8= (.7) (7)^n

    hydride do you have fred at csuf? lol
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