Help with Exponential Equations, Solving y=ab^x?

Question

I have a table of data:

x y
2 -2
4 -8
6 -32

How can I solve this. Thanks.

Hi

Idest23, it wasn't quite obvious to me which is why I asked for help.

Quilasheila, thanks for helping me work the problem. That answers my question and helps me to learn to do the problem

Yes they were negatives. Thanks.

Answer 1

There are two unknowns.
Create 2 equations to solve

Plug in (2,2) for the first equation
2 = a * b^2

now for second equation plug in (4,8)
8 = a * b^4

We have two equations with two unknowns.
Divide eqn two by eqn one

8 = a * b^4
_________ ===> now the a's cancel
2 = a * b^2

4 = b^2
b = 2

Now plug b = 2 into any of the equations to get a
(I will use the first equation)
2 = a * b^2
2 = a * (2^2)
2 = 4a
a = 1/2

So A=1/2 and B=2

y = 1/2 * 2^x


=]
_____________________________________
oops! Are those neagatives? or separating the x and y?

If you meant negative, then do above with different points.
(2, -2) ===> -2 = a * b^2
(4, -8) ===> -8 = a * b^4
divide again -----> 4 = b^2
b = 2
plug in b to get a
-2 = a * b^2
-2 = a *2^2
4a = -2
a = -1/2

y = -1/2 * 2^x

which also equals y = - ( 2^(x-1) )
=]


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..

Answer 2

nicely, 9 = ?3^4 so which you will rewrite the appropriate facet of the equation as ?3^4(x-2). because the two factors now have the comparable base, you could basically set the exponents equivalent to a minimum of one yet another. 2x+4 = 4(x-2) 2x+4 = 4x-8 2x = 12 x = 6

Answer 3

Well its quite obvious that a = -1/2 and b = 2.

Answer 4

I'm not sure how you would solve this because your formula would be in terms of a and b. Without knowing what a or b is, you are going to get stuck.

Now, the question is, is (ab)^x or is a * b^x.

First one:

y^(-x) = ab

Second one:

y/a = b^x
(y/a)^(-x) = b

Answer 5

for x=2 and y=2
equation is 2=ab^2
taking sq root both sides we get
ab=sq root of 2

for x=4 and y=8
8=ab^4
so ab =fourth root of 8

for x=6 and y=32
32=ab^6
ab=sixth root of 32

Answer 6

It is not clear exactly what you want to solve for.