Exponential Graph Rule Problem

[chiara__]

Hi! For my school assignment I received this worded question using exponential graphs, but I am having some trouble with the first part about finding the rule for it.

Since it is an exponential graph, I determined the rule used would be y=a^-x, and then tried to go from there. Since 15000 is the initial value I placed that in the rule, and I also substituted the point (0,15000) since I know that will be on the graph. I think I have done this right but I ended up with an extremely long answer of


V = 15000 * (2.8045882272237)^-0.06*x

Does this seem correct to anyone? The only reason I am thrown off is because of the length of the base of -0.06x; because I have never seen an equation with a number with so many significant figures... especially not on our usual school assignments. I tried rounding it, but that changes the answer, so I'm stuck with this super long answer that I'm not even sure is right...

Does anyone think this is completely wrong? Or is the very long answer right, and I am just overthinking it?

Thank you so much for any help you are able to provide

(I've attached the question below)

 

[Deleted member 4993]

Hi! For my school assignment I received this worded question using exponential graphs, but I am having some trouble with the first part about finding the rule for it.

Since it is an exponential graph, I determined the rule used would be y=a^-x, and then tried to go from there. Since 15000 is the initial value I placed that in the rule, and I also substituted the point (0,15000) since I know that will be on the graph. I think I have done this right but I ended up with an extremely long answer of


V = 15000 * (2.8045882272237)^-0.06*x

Does this seem correct to anyone? The only reason I am thrown off is because of the length of the base of -0.06x; because I have never seen an equation with a number with so many significant figures... especially not on our usual school assignments. I tried rounding it, but that changes the answer, so I'm stuck with this super long answer that I'm not even sure is right...

Does anyone think this is completely wrong? Or is the very long answer right, and I am just overthinking it?

Thank you so much for any help you are able to provide

(I've attached the question below)

Can you please show us - how you derived the equation?

Start with an equation expression of volume as a function of time [V(t) = ?]

 

[Dr.Peterson]

Hi! For my school assignment I received this worded question using exponential graphs, but I am having some trouble with the first part about finding the rule for it.

Since it is an exponential graph, I determined the rule used would be y=a^-x, and then tried to go from there. Since 15000 is the initial value I placed that in the rule, and I also substituted the point (0,15000) since I know that will be on the graph. I think I have done this right but I ended up with an extremely long answer of


V = 15000 * (2.8045882272237)^-0.06*x

Does this seem correct to anyone? The only reason I am thrown off is because of the length of the base of -0.06x; because I have never seen an equation with a number with so many significant figures... especially not on our usual school assignments. I tried rounding it, but that changes the answer, so I'm stuck with this super long answer that I'm not even sure is right...

One good thing to do is to calculate directly (just from the problem statement, without a formula) how much is left after the first hour. Then check whether your formula gives that value.

It is quite possible that in doing this you will get a better idea what the formula should look like -- in particular, where the 6% belongs.

 

[jonah2.0]

Beer soaked hint follows.

Hi! For my school assignment I received this worded question using exponential graphs, but I am having some trouble with the first part about finding the rule for it.

Since it is an exponential graph, I determined the rule used would be y=a^-x, and then tried to go from there. Since 15000 is the initial value I placed that in the rule, and I also substituted the point (0,15000) since I know that will be on the graph. I think I have done this right but I ended up with an extremely long answer of


V = 15000 * (2.8045882272237)^-0.06*x

Does this seem correct to anyone? The only reason I am thrown off is because of the length of the base of -0.06x; because I have never seen an equation with a number with so many significant figures... especially not on our usual school assignments. I tried rounding it, but that changes the answer, so I'm stuck with this super long answer that I'm not even sure is right...

Does anyone think this is completely wrong? Or is the very long answer right, and I am just overthinking it?

Thank you so much for any help you are able to provide

(I've attached the question below)

Geometric Progressions.

 

[Dr.Peterson]

V = 15000 * (2.8045882272237)^-0.06*x

Does this seem correct to anyone? The only reason I am thrown off is because of the length of the base of -0.06x; because I have never seen an equation with a number with so many significant figures... especially not on our usual school assignments. I tried rounding it, but that changes the answer, so I'm stuck with this super long answer that I'm not even sure is right...

If you have followed my suggestion, you will have found that your formula is actually correct. But it is not written in the most reasonable form, which I suspect is a result of mixing up two ways of thinking about exponential functions, namely the form [MATH]e^{kt}[/MATH] and [MATH]a^t[/MATH]. Your form is [MATH]a^{kt}[/MATH]! And that is what makes it overly complicated.

Try rewriting [MATH]2.8045882272237^{-0.06x}[/MATH] as [MATH]\left(2.8045882272237^{-0.06}\right)^{x}[/MATH] and simplify it, and you'll find a much nicer formula!

This is where SK's request is important. If you show us your thinking, we can help you change it to a quicker and much easier way.

 

[jonah2.0]

Beer inspired realization follows.

It just dawned on me that the premise of this problem is somehow unrealistic.
A leak at the bottom of the tank would cause a faster water drain in the first few hours.
To have a leak that extracts 6% of the water remaining in a tank every hour is another unrealistic scenario. Whoever composed this problem must have been drinking some really good stuff. It would have been better to have given this problem as a depreciation scenario.

 

[skeeter]

 

[chiara__]

Can you please show us - how you derived the equation?

Start with an equation expression of volume as a function of time [V(t) = ?]

So, the equation I started with was [V(t) = V₀ * a^-k*t

Then, I input 15000 as V₀ since it is the initial amount, and also replaced the k with 0.06, as that was the only other variable I had.

This left me with [V(t) = 15000 * a^-0.06t . From here, I knew that (0, 15000) was a point so I put that into the equation and solved for 'a' in my calculator, which is how I came to the lengthy A= 2.8045882272237.

Therefore my final equation ended up as [V(t) = 15000 * (2.8045882272237)^-0.06t .

Prior to even figuring out the equation, I spent some time working out other points, for t=1,2,3,etc without the formula, simply subtracting 6% from each individual value, which is how I was able to check that only this formula, with the incredibly long number, gave me the exact answers I needed.

 

[chiara__]

If you have followed my suggestion, you will have found that your formula is actually correct. But it is not written in the most reasonable form, which I suspect is a result of mixing up two ways of thinking about exponential functions, namely the form [MATH]e^{kt}[/MATH] and [MATH]a^t[/MATH]. Your form is [MATH]a^{kt}[/MATH]! And that is what makes it overly complicated.

Try rewriting [MATH]2.8045882272237^{-0.06x}[/MATH] as [MATH]\left(2.8045882272237^{-0.06}\right)^{x}[/MATH] and simplify it, and you'll find a much nicer formula!

This is where SK's request is important. If you show us your thinking, we can help you change it to a quicker and much easier way.

This helps so much, thank you! I did calculate the values for t=1,2,3,etc without the equation at first, just to have answers for the other parts of the question in case I couldn't figure out the equation, but to be honest that didn't really help me in figuring out where to place the 6%. All it did was prove that my equation, although lengthy, worked with the values I had previously calculated.

I've now tried placing the -0.06 inside the bracket with the 2.8045882272237, instead of outside with the 't', and that was so helpful with finally getting the answer down to a much simpler and nicer equation, so thank you

I'm just wondering though, how am I supposed to know to place the 6% inside the bracket rather than out? That's the one thing I am still not really understanding... I just figured it had to be on the outside, but that's obviously not the case. Is there anything that you think could help me understand this?

 

[jonah2.0]

Beer hangover vision follows.

So, the equation I started with was [V(t) = V₀ * a^-k*t

Then, I input 15000 as V₀ since it is the initial amount, and also replaced the k with 0.06, as that was the only other variable I had.

This left me with [V(t) = 15000 * a^-0.06t . From here, I knew that (0, 15000) was a point so I put that into the equation and solved for 'a' in my calculator, which is how I came to the lengthy A= 2.8045882272237.

Therefore my final equation ended up as [V(t) = 15000 * (2.8045882272237)^-0.06t .

Prior to even figuring out the equation, I spent some time working out other points, for t=1,2,3,etc without the formula, simply subtracting 6% from each individual value, which is how I was able to check that only this formula, with the incredibly long number, gave me the exact answers I needed.

This helps so much, thank you! I did calculate the values for t=1,2,3,etc without the equation at first, just to have answers for the other parts of the question in case I couldn't figure out the equation, but to be honest that didn't really help me in figuring out where to place the 6%. All it did was prove that my equation, although lengthy, worked with the values I had previously calculated.

I've now tried placing the -0.06 inside the bracket with the 2.8045882272237, instead of outside with the 't', and that was so helpful with finally getting the answer down to a much simpler and nicer equation, so thank you

I'm just wondering though, how am I supposed to know to place the 6% inside the bracket rather than out? That's the one thing I am still not really understanding... I just figured it had to be on the outside, but that's obviously not the case. Is there anything that you think could help me understand this?

Nice work.
Common type of problems with depreciation stuff.
I've seen my good friend, the late indefatigable Math Knight-Errant Sir Denis, solve this stuff like clockwork.
Equivalently, you could also simply write [MATH]15000(1-.06)^t[/MATH] or [MATH]15000(.94)^t[/MATH]

 

[Dr.Peterson]

So, the equation I started with was [V(t) = V₀ * a^-k*t

Then, I input 15000 as V₀ since it is the initial amount, and also replaced the k with 0.06, as that was the only other variable I had.

This left me with [V(t) = 15000 * a^-0.06t . From here, I knew that (0, 15000) was a point so I put that into the equation and solved for 'a' in my calculator, which is how I came to the lengthy A= 2.8045882272237.

Therefore my final equation ended up as [V(t) = 15000 * (2.8045882272237)^-0.06t .

I've now tried placing the -0.06 inside the bracket with the 2.8045882272237, instead of outside with the 't', and that was so helpful with finally getting the answer down to a much simpler and nicer equation, so thank you

I'm just wondering though, how am I supposed to know to place the 6% inside the bracket rather than out? That's the one thing I am still not really understanding... I just figured it had to be on the outside, but that's obviously not the case. Is there anything that you think could help me understand this?

The problem is, as I indicated, that you don't need both [MATH]a[/MATH] and [MATH]k[/MATH] in your general equation. There are several forms commonly taught, and your [MATH]V_0a^{-kt}[/MATH] is not one of them, in my experience. You need only one calculated constant, either your [MATH]a[/MATH] or your [MATH]k[/MATH]; the formula might take any of these forms, depending on your specific needs:

[MATH]V_0e^{-kt}[/MATH]​

[MATH]V_010^{-kt}[/MATH]​

[MATH]V_02^{-kt}[/MATH]​

[MATH]V_0a^{t}[/MATH]​

[MATH]V_0(1+r)^{t}[/MATH]​

That is, you can either arbitrarily choose the base (which might be [MATH]e[/MATH], [MATH]10[/MATH], or [MATH]2[/MATH]), and use the data to determine the coefficient in the exponent; or you can make the exponent be [MATH]t[/MATH] alone and use the data to determine the base, either as a number [MATH]a[/MATH] or as [MATH]1+r[/MATH], which I'll explain in a moment. (There are a couple other variations if you know the half-life or doubling time, or other special cases.)

In a case where you are given the percent change per unit time, [MATH]r[/MATH], you know that the quantity is multiplied by [MATH]1+r[/MATH] each unit; so that is the base! Therefore, using the last form I listed, you can immediately write down the formula for your problem by filling in [MATH]V_0 = 15000[/MATH], and [MATH]r = -0.06[/MATH].

When you simplified the form I showed, you found the base to be approximately 0.94. If you do it this way, you immediately find that [MATH]1+r = 1 - 0.06 = 0.94[/MATH], exactly. That is, each hour the volume is multiplied by 0.94, which is 1.00 - 0.06, a reduction of 6%. And that is how you know what to put there. I'm very surprised that you weren't taught it in this form.