## anonymous one year ago How to find the variables a and b, in the function y = ax^b, using algebraic methods?

1. anonymous

I already have a table of values for x and y. I need to use an algebraic method other than linear regression to model the function.

2. myininaya

use the points to find a and b for example say you have (1,4) and (2,16) on your graph then we have $4=a(1)^b \text{ and } 16=a(2)^b \\ \text{ so that is we have } \\ 4=a \text{ and } 16=4(2)^b \\ \text{ so now we only need \to solve } \\ 16=4(2)^b \\ 4=2^b \\ \text{ then we know }2^2=4 \text{ so } b=2 \\ \text{ so if } (1,4) \text{ and } (2,16) \text{ are on the graph } \\ \text{ the our function looks like } f(x)=4x^2$

3. anonymous

Okay. For example, how would I find a and b in the following functions: 0.000249 = a280b 0.000122 = a180b

4. anonymous

It is to the power of b.

5. myininaya

$2.49 \cdot 10^{-4}=a(280)^b \\ 1.22 \cdot 10^{-4}=a(180)^b \\ \text{ you could solve the first equation for } a \text{ then replace the } a \\ \text{ in the second equation with what you found it to be from the first } \\ \text{ this is what I did in the example \above }$ In general say we the points (c,d) and (m,n) $d=a(c)^b \text{ and } n=a(m)^b \\\text{ solve first for } a \\ d(c)^{-b}=a \\ \text{ so now we have \in the second equation } \\ n=d(c)^{-b}(m)^b \text{ by law of exponents you have } \\ n=d(c^{-1}m)^b \\ \text{ now we are trying \to solve for } b \\ \text{ divide both sides by} d \\ \frac{n}{d}=(c^{-1}m)^b \\ \text{ take } \ln( ) \text{ of both sides } \\ \ln(\frac{n}{d})=b \ln(c^{-1}m) \\ \text{ now multiply both sides by } \frac{1}{\ln(c^{-1}m)} \\ \\ \frac{1}{\ln(c^{-1}m)}\ln(\frac{n}{d})=b \\ \text{ so we have found } b \\ \text{ now we can go back \in find } a \\ \text{ recall } a=d(c)^{-b} \\ a=d(c)^{-\frac{1}{\ln(c^{-1}m)} \ln(\frac{n}{d})}$

6. myininaya

so try solving your first equation for a

7. myininaya

then pluggin that into the second equation

8. myininaya

please let me know if you still don't follow

9. anonymous

Will this be as accurate as linear regression, considering there are several values?

10. myininaya

I don't know. I would have to have the set of all the points. And do a comparison with this method (whatever method this is called) versus the linear regression method Basically I would want to see if more points fall closer to the line or the the thing that we find from doing f=ax^b

11. myininaya

by the way the value you get for a is really close to 0 using these points you chose but anyways I think I definitely would suggest writing it in scientific notation otherwise you will have a lot of zeros to write :p