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#1
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Reversing POWER Formula
All,
Could someone with greater algebra skills help me with below formula? Cell F2 contains 0.5461% Cell G2 contains =POWER(F2+1,12)-1, which equals 6.7542% I am trying to create a formula that reverses the above calculation. I would receive cell G2 data from a system (6.7542%)and would need to calculate in Excel non-annualized data in cell F2 (0.5461). I remember tackling something like this many, many years ago but can't quite summon the required math (perhaps a logarithm)? Thanks in advance for any ideas. Regards, Joel Using Excel 2003 |
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#3
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Hi Joel,
This works for me =(G2+1)^(1/12)-1 -- HTH RP (remove nothere from the email address if mailing direct) wrote in message ups.com... All, Could someone with greater algebra skills help me with below formula? Cell F2 contains 0.5461% Cell G2 contains =POWER(F2+1,12)-1, which equals 6.7542% I am trying to create a formula that reverses the above calculation. I would receive cell G2 data from a system (6.7542%)and would need to calculate in Excel non-annualized data in cell F2 (0.5461). I remember tackling something like this many, many years ago but can't quite summon the required math (perhaps a logarithm)? Thanks in advance for any ideas. Regards, Joel Using Excel 2003 |
#4
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Just to be different...
=NOMINAL(G2,12)/12 Are you trying to calculate effective rate by chance in G2? HTH :) -- Dana DeLouis Win XP & Office 2003 wrote in message ups.com... All, Could someone with greater algebra skills help me with below formula? Cell F2 contains 0.5461% Cell G2 contains =POWER(F2+1,12)-1, which equals 6.7542% I am trying to create a formula that reverses the above calculation. I would receive cell G2 data from a system (6.7542%)and would need to calculate in Excel non-annualized data in cell F2 (0.5461). I remember tackling something like this many, many years ago but can't quite summon the required math (perhaps a logarithm)? Thanks in advance for any ideas. Regards, Joel Using Excel 2003 |
#6
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You didn't answer Dana's question. I would be interested to know as well.
-- HTH RP (remove nothere from the email address if mailing direct) wrote in message oups.com... Thanks for above responses. This is exactly what I needed! |
#7
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Hi. What we're trying to say is that your equation may be correct. We just
want to point out that there seems to be a big difference from entering 0.5461% and returning 6.7542%. If you changed your equation slightly to include the "/12" part... =POWER(0.5461%+1,12)-1 to this.. =POWER(0.5461%/12+1,12)-1 You would get the same answer as =EFFECT(0.5461%,12) which would be 0.5475%. Again, just thought we would mention it. :) -- Dana DeLouis Win XP & Office 2003 "Bob Phillips" wrote in message ... You didn't answer Dana's question. I would be interested to know as well. -- HTH RP (remove nothere from the email address if mailing direct) wrote in message oups.com... Thanks for above responses. This is exactly what I needed! |
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