# Power Functions

• Nov 26th 2009, 08:39 PM
hydride
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!
• Nov 26th 2009, 09:08 PM
CaptainBlack
Quote:

Originally Posted by 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:
$8 = k7^p$
$0.7 = k1^p$

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
• Nov 26th 2009, 09:34 PM
Soroban
Hello, hydride!

Quote:

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

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

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

• Nov 26th 2009, 09:39 PM
hydride
How did you get k = .7?

Can you do it an alternate way? By using substitution?
• Nov 27th 2009, 03:54 AM
HallsofIvy
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 $p= \frac{log(11.42857)}{log(7)}$.
• Nov 27th 2009, 08:07 AM
hydride
Quote:

Originally Posted by HallsofIvy
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?
• Nov 28th 2009, 11:02 PM
getevlcted
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