## Recommended Posts

Hello forum.  Trying to modify a formula to solve for a different variable.  Consider this formula which I am using and works fine:

x*(y/100)^.5

Here are the fields in the database.
Round(PUMPPRESSURE::gNozzSize*(PUMPPRESSURE::gOldPress/100)^.5;2)

In this case when
x=.75
y=145
The answer is rounded to .90  Works fine.

The first part of my question is this...
What I am trying to figure out is a formula that uses the previous result of .90, and if I change x to .65 get the new formula to provide an answer that calculates for y.

With the above numbers, when I change x to .65 in this new calculation, solving for y should equal 190.  I can back into it with the first formula by consistently changing y until I get the .90 answer.

Just looking to automate it.

I'm usually pretty good with this but can't wrap my head around it.  The exponent .5,  is making my head hurt, LOL.
Thanks

Steve

Edited by Steve Martino
##### Share on other sites

I am not sure I understand the question. Let's say the formula is:

z = x * ( y / 100)^.5

so that when the input is:
x = .75 ;
y = 145

then the output will be:

z = .9031195934094222

Are you asking how to calculate the value of y from the input of x and z? If so, the answer will be:

y = 100 * ( z / x )^2

Edited by comment
##### Share on other sites

You understood it perfectly.  That is correct.  Thanks!

##### Share on other sites

Ah, good. Just for future reference:

n^.5 is the same thing as Sqrt(n) .  So the inverse of y = x^.5 is x = y^2.

• 1
##### Share on other sites

Thanks.  I last had calculus (and had any reason to use it!) in high school, in 1983, so much lost and forgotten.  Thanks again.

## Create an account

Register a new account