Geomean function
BobA5835 wrote...
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Interestingly enough, your formula gave me a closer value to the
target test value than Harlans. My programmer says it may have to
do with the log function in some way.
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It might. If this were so, you might achieve even higher precision
using a larger base for your logarithms, e.g.,
=128^AVERAGE(LOG(x,128))
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