Well, it helps to start out with the right formula. Instead of

DECIBEL = V1/V2 LOG 20

the formula is actually

DECIBEL = 20 * LOG(V1/V2)

So, just rearranging and taking the antilog

V2 = V1/10^(DECIBEL/20)



In article ,
Johno wrote:

Thanks but I'm not so sure this works out exactly, if I use 1OO as V1
and 50 as V2 then the decibel figure is 6.02 .... if I then work back
using the formula V2 = V1/Decibel Log 20 instead of V2 equalling 50 as
in the first formula it equals 6.41 not 6.02 and if other figures are
substituted and then worked back the error varies .... can anyone shed
some light. ... Regards