How to find the variables a and b, in the function y = ax^b, using algebraic methods?

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

How to find the variables a and b, in the function y = ax^b, using algebraic methods?

- chestercat

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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.

- 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
\[4=a(1)^b \text{ and } 16=a(2)^b \\ \text{ so that is we have } \\ 4=a \text{ and } 16=4(2)^b \\ \text{ so now we only need \to solve } \\ 16=4(2)^b \\ 4=2^b \\ \text{ then we know }2^2=4 \text{ so } b=2 \\ \text{ so if } (1,4) \text{ and } (2,16) \text{ are on the graph } \\ \text{ the our function looks like } f(x)=4x^2\]

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

It is to the power of b.

- 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)
\[d=a(c)^b \text{ and } n=a(m)^b \\\text{ solve first for } a \\ d(c)^{-b}=a \\ \text{ so now we have \in the second equation } \\ n=d(c)^{-b}(m)^b \text{ by law of exponents you have } \\ n=d(c^{-1}m)^b \\ \text{ now we are trying \to solve for } b \\ \text{ divide both sides by} d \\ \frac{n}{d}=(c^{-1}m)^b \\ \text{ take } \ln( ) \text{ of both sides } \\ \ln(\frac{n}{d})=b \ln(c^{-1}m) \\ \text{ now multiply both sides by } \frac{1}{\ln(c^{-1}m)} \\ \\ \frac{1}{\ln(c^{-1}m)}\ln(\frac{n}{d})=b \\ \text{ so we have found } b \\ \text{ now we can go back \in find } a \\ \text{ recall } a=d(c)^{-b} \\ a=d(c)^{-\frac{1}{\ln(c^{-1}m)} \ln(\frac{n}{d})}\]

- myininaya

so try solving your first equation for a

- myininaya

then pluggin that into the second equation

- myininaya

please let me know if you still don't follow

- anonymous

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

- 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

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.