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#### goosey00

##### Member
The table shows a purchasing power of a dollar of various years. Use the first and last data points to find an exponential function, and use it to determine when the purchasing power of a dollar will drop to 40 cents. Let t=0 CORRESPOND TO 1983-top being year an bottom being purchasing power
 1983 1985 1987 1989 1991 1993 1995 1.02 0.98 0.95 0.91 0.88 0.85 0.82
The purchasing power will reach 40 cents in what year??

I get this whole problem all the way up to the part where you have to find the purchasing power is 40 cents. To do this i'm suppose to get it by, graph y=(t) and y=.40 find where they intersect. It says to round to the nearest who number. How do they even intersect? I see them as being parallel.?? Help please!!!

#### Jameson

Staff member
What you need to do is first find a way to model the decay of purchasing power of the dollar. That will be a function of time, t. Then you need to find out for what t, f(t)=.4. Did you already find the modeling function?

#### goosey00

##### Member
yes
I am stuck at the last part that says to check by graphing

#### Jameson

Staff member
yes
I am stuck at the last part that says to check by graphing
What did you get for your equation? It's a tiny bit tricky actually since you have this pattern of decreasing by 4,3,4,3,4,3. What year did you find that f(t)=.4?

Also, the problem says this is exponential but it's not decaying exponentially. It's decaying linearly. That's a bit strange.

#### goosey00

##### Member
Ok-so here is wher I got to it and now Im stuck-
1.02=.82
then b^12=.80392
b=.981976
Then it says wtite the values of y(small under zero)0+b
Then it says 1.02(.98176)^t

#### goosey00

##### Member
The part that gives the years is the odd graphing part. Thats where im stuck.

#### Jameson

Staff member
Ok-so here is wher I got to it and now Im stuck-
1.02=.82
then b^12=.80392
b=.981976
Then it says wtite the values of y(small under zero)0+b
Then it says 1.02(.98176)^t
Ok, I guess exponential modeling does work. We start with $$\displaystyle P=a \cdot b^t$$ where P is the new purchasing power for t years after 1983, a is the starting purchasing power, b is the growth or decay rate and t the number of years after 1983.

That gives us $$\displaystyle P=1.02 \cdot b^t$$

Now we need to solve for b. Let's use the year 1985 and purchasing power .98.

$$\displaystyle .98 = 1.02 \cdot a^2$$

Solving for "b" we get b = 0.9802. If you do the same calculation with a different year, say 1991 and .88 you get $$\displaystyle .88 = 1.02 \cdot a^8$$ and that b=0.9817.

Finally if you use the last data point you get $$\displaystyle .82 = 1.02 \cdot b^{12}$$ and b=0.981977 So I don't see where you got "b=0.98176" from but all of these calculations have similar values.

Anyway, graphing it. Although I don't know where you got "b=0.98176" from the graph would look like this.

$$\displaystyle P=1.02(.98176)^t$$

[graph]xvknk86mgj[/graph]

It looks almost like a line. What did you get for the year when P(t)=.4?

EDIT: I think I see what you did. When I solve $$\displaystyle .82 = 1.02 \cdot b^{12}$$ on my calculator I get b=.981976. Did you forget to write a 9?

That graph looks almost identical anyway, as you can see below.

[graph]2lolxw1tcr[/graph]

EDIT2: Ok, I see that you just missed that 9 when you wrote it a second time. Now we're on the same page. Were you just having trouble graphing it or having trouble showing how the graph confirms your answer?

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#### goosey00

##### Member
1.02b^14=.40
b^12=..80392^(1/14)
Then I got .981976

1.02(.98176)^t
To answer it, yes. I am having trouble graphing it. The above is the last part I am stuck

#### goosey00

##### Member
will you help me on how to graph even the one you show. What did you enter?

#### Jameson

Staff member
1.02b^14=.40
b^12=..80392^(1/14)
Then I got .981976

1.02(.98176)^t
To answer it, yes. I am having trouble graphing it. The above is the last part I am stuck
How did you find t=14? If you plug this into Mathematica you get this . Your answer isn't correct but your question I think is how to graph.

Do you want to graph it online or on your calculator?

1.02 * (.981976) ^ x

or

1.02 * (.981976) (x^y) x

Note: * means "times".

EDIT: To see how I included the graphs in my above post, I made a tutorial video that you can find in this thread.

EDIT2: Also, if you want to see how I graphed the below graph you can just click on it. It's a link.

[graph]2lolxw1tcr[/graph]

Last edited:

#### goosey00

##### Member
Got the one equation perfect. What do you then input for .4? Is there also an equation. I see ho you did it and it makes sense but the graphing calculator doesn't seem to want to plug it in as just .4 and intersect.

#### Jameson

Staff member
Got the one equation perfect. What do you then input for .4? Is there also an equation. I see ho you did it and it makes sense but the graphing calculator doesn't seem to want to plug it in as just .4 and intersect.
Here's how you can do it on Desmos.

[graph]6fw9uatd65[/graph]

What graphing calculator do you have? You need to graph two equations.

y = 1.02*(.981976^x) and y = .4

#### goosey00

##### Member
Im just using this one right now. Obviously I have to get a "real" one tomorrow-huh?? Graphing Calculator

#### Jameson

Staff member
Im just using this one right now. Obviously I have to get a "real" one tomorrow-huh?? Graphing Calculator
I don't know what you need to get for your class, since I have no idea what your teacher or professor has told you . Do you understand how to graph both equations though? This thread is getting kind of long and I think I've misinterpreted what you want help with. It seems you just want to know how to graph things.

Is everything more or less clear on how to graph everything?

#### goosey00

##### Member
just graphing-I know-you should be getting paid for this one. I am sooo sorry.
With the calculator I showed you, where would I put in .4