On Feb 15, 6:54*pm, Shawn wrote:
Here is the formula set up to solve for MGD (Million Gallons per Day).
=3*936*A21^1.5*1440/1000000
936 is the length in inches.
A21 is a cell with the inches going over the weir.
1440/1000000 is to convert the gallons per minute to MGD.
What I want to do is set up another formula where the desired MGD
could be entered in a cell and the result would be the inches to let
flow over the weir.

I know nothing of the application, but is this simple algebra? If,
say, B21 is the MGD and you want A21 to be the inches going over the
weir, then:

A21^1.5 = B21*1000000/3/936/1440

log(A21^1.5) = log(B21*1000000/3/936/1440)

1.5*log(A21) = log(B21) + log(1000000) - log(3) - log(936) - log(1440)

log(A21) = (log(B21) + log(1000000) - log(3) - log(936) - log(1440)) /
1.5

A21: =10^((log(B21) + log(1000000) - log(3) - log(936) - log(1440)) /
1.5)

Check: In C21, write:

=3*936*A21^1.5*1440/1000000

Should approximately equal B21. Don't be surprised by some small
difference, the result of numerical approximations of the log's as
well as binary computer arithmetic anomalies. (I got an error of
about -7E-14.)