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It's been too long since I've had to do this kind of math, so I'm hoping to get some help. I need to figure out the $x$ to $y$ relationship so that I can properly calculate the capacity of some filters. On the left side of the equation I have flow rate and on the right side I have the amount of water that can be treated at that flow rate. Every time the flow is doubled the amount that can be treated goes down by a factor of $2.5$. The opposite is also true, every time the flow is halved the amount goes up by a factor of $2.5$.

Here's the data I'm working with: $2.5$ gpm = $25000$ gallons and $5$ gpm = $10000$ gallons

How can I solve the amount for the flow rate being $1$ gpm, or $7$ gpm, etc.

Thanks

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    By "goes up by" do you mean that it increases by *adding* 2.5, or by *multiplying* 2.5? Based on your data I assume you mean the latter. – 2017-01-06
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    That's correct it's the latter, it's multiplied by 2.5. – 2017-01-06

1 Answers 1

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Relationships like this can be modeled by a power function of the form $y = ax^r$ where $a$ and $r$ are constants. In particular, substituting $x=2.5,y=25000$ we have $$25000=a(2.5)^r$$ and substituting $x=5,y=10000$ we have $$10000=a5^r$$ If we divide the first equation by the second, we have $$2.5=(0.5)^r$$ which has the solution $r=\frac{\log 2.5}{\log 0.5} \approx -1.322$. Back-substituting this into the first equation, we have $$25000 = a (2.5)^{-1.322}$$ which has the solution $a \approx 83949$. Putting this all together, we have $$y = 83949x^{-1.322}$$ as a reasonable model for your data. (Plug in $x=2.5$, $x=5$, and $x=10$ to see what the corresponding values of $y$ are.)

Now just plug in any value of $x$ you want to see what the corresponding value of $y$ is.

If you want a more exact formula (one that doesn't rely on decimal approximations) you can proceed as follows: Back-substitute the exact value $r=\frac{\log 2.5}{\log 0.5}$ into the first equation to get $$25000 = a (2.5)^{\frac{\log 2.5}{\log 0.5}}$$ the solution of which is $$a=25000(2.5)^{-\frac{\log 2.5}{\log 0.5}}$$ so that the final equation is $$y=25000(2.5)^{-\frac{\log 2.5}{\log 0.5}} x^{\frac{\log 2.5}{\log 0.5}}$$ which can be simplified to $$y=25000\left(\frac{x}{2.5}\right)^{\frac{\log 2.5}{\log 0.5}}$$