POWER function using DATEDIFF

[heardm]

I need to raise a value in one column to the date diff of two other columns. I calculated the DATEDIFF in a separate calculated column. All of the column values are around 4. Then I tried to make a new calculated column that raised column values to the power of that date difference, but I get this error

"An argument of function 'POWER' has the wrong data type or the result is too large or too small."

All of my columns have the data type "decimal number"

[heardm · Accepted solution]

The problem was that I had previously divided two columns and if a row had a missing value I was given an answer of infinity, which I then tried to use in the POWER function

[v-jiascu-msft]

Hi @heardm

Could you please mark the proper answer if it's convenient for you? That will be a help to others.

Best Regards!
Dale

[Chihiro]

That error can be caused by having too large of a number in 2nd argument of POWER function as well.

See image for an example.