1. ## Exponential Functions

I can't figure out how to find the base of the exponential functions...really ticking me off lol. I'd appreciate any help. I have two different problems

Problem 1:

Determine growth or decay factor

x |-2 | -1 | 0 | 1 | 2 | 3 | 4
___________________________
y|6.25|2.5 | 1 | .4 | .16|.064| .0256

the decimals really screw it up, and I've tried many numbers...is there any kind of formula to figure this out? our teacher just told us that we might have to do some "guess work"

Problem 2:

Assume that y is an exponential function of x

If the growth factor is 1.08, then complete the following table

x 0 | 1 | 2 | 3
_____________________
y 23.1|

They gave me the first Y point... I can't figure out any of the others...I'm pretty bad at math...if anybody knows a formula I'd greatly appreciate it.

2. Problem 1:

Well lets examine these numbers.
$1 \rightarrow .4,$
$2 \rightarrow .16,$
$3 \rightarrow .064...$
What do you see here? What is
$.4^1,$
$.4^2,$
$.4^3?$

The answer is $y(x)= .4^x$

3. Thanks! I don't know how I didn't see that in the first problem...guess I didn't look hard enough...

Second problem, I don't understand how y is 23.1 but x is 0, and using that with the information you gave me..if you could explain a bit more, I'd appreciate it tons..

4. Hello, ohiostatefan!

1) Determine growth or decay factor.

. . $\begin{array}{c||c|c|c|c|c|c|c|}\
x & \text{-}2 & \text{-}1 & 0 & 1 & 2 & 3 & 4 \\ \hline
y & 6.25& 2.5 & 1 & 0.4 & 0.1& 0.064 & 0.0256 \end{array}$
There is no guesswork . . .

You're expect to know the general exponential function: . $f(x) \:=\:a\cdot b^x$

Use two values from the table to determine $a$ and $b.$

. . $f(0) = 1\!:\;\;a\cdot b^0 \:=\:1 \quad\Rightarrow\quad a \:=\:1$

. . $f(1) = 0.4\!:\;\;1\cdot b^1 \:=\:0.4 \quad\Rightarrow\quad b \:=\:0.4$

The function is: . $f(x) \:=\:(0.4)^x$

The decay factor is: . $b \,=\,0.4$

2) Assume that y is an exponential function of x.

If the growth factor is 1.08, then complete the following table

. . $\begin{array}{c||c|c|c|c}
x & 0 & 1 & 2 & 3 \\ \hline
y & 23.1 & & & \end{array}$

Start with: . $f(x) \:=\:a\cdot b^x$

We are given: . $b \,=\,1.08$

So the function is: . $f(x) \:=\:a(1.08)^x$

We are given: . $f(0) = 23.1$

So we have: . $23.1 \:=\:a(1.08)^0 \quad\Rightarrow\quad a\:=\:23.1$

. . And the function is: . $f(x) \:=\:23.1(1.08)^x$

Therefore:

. . $f(1) \;=\;2.31(1.08)^1 \;=\;2.4948$

. . $f(2) \;=\;2.31(1.08)^2 \;=\;2.694384$

. . $f(3) \;=\;2.31(1.08)^3 \;=\;2.90993472$

5. Thank you so much for the help! I understand it now... the only part I don't understand is on problem 2, how the 23.1 turns into 2.31. I've been looking it over and I can't figure it out. I know you've helped me alot already...so if you don't feel like it, I understand :P

6. I believe it is a typo.

Use $f(x)= 23.1(1.08)^x$

i.e.,

$f(0)= 23.1(1.08)^0$
$f(1)= 23.1(1.08)^1$

$...$

Or, move all the decimals of the posted answers one place to the right.

Originally Posted by Soroban

Therefore:

. . $f(1) \;=\;23.1(1.08)^1 \;=\;24.948$

. . $f(2) \;=\;23.1(1.08)^2 \;=\;26.94384$

. . $f(3) \;=\;23.1(1.08)^3 \;=\;29.0993472$

7. Thought it was a typo, just wasn't sure if I was doing something wrong, lol. Thanks, Anonymous1.