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Math Help - Power Functions

  1. #1
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    Power Functions

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

    This is what I've gotten only:
    8 = k7^p
    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)
    f(x) = kx^p

    This is what I've gotten only:
    8 = k7^p
    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 y=kx^p becomes:

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

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

    f(x) = kx^p

    This is what I've gotten only:
    8 \:=\: k\!\cdot\!7^p
    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: . \begin{array}{cccc}7^p\!\cdot\!k &=& 8 & [1] \\ k &=& 0.7 & [2] \end{array}

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

    Take logs, base 7: . \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: . p \:=\:\log_7(\tfrac{80}{7})


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


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


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

    *
    \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 8= k 7^p and .7= k 1^p= k we have immediately that k= .7. Put that into the first equation: 8= .7(7^p) so 7^p= 8/.7= 11.42857, approximately. Now take the logarithm of boht sides: p log(7)= log(11.42857) and 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 8= k 7^p and .7= k 1^p= k we have immediately that k= .7. Put that into the first equation: 8= .7(7^p) so 7^p= 8/.7= 11.42857, approximately. Now take the logarithm of boht sides: p log(7)= log(11.42857) and 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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