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rearrange to isolate A
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\[H = K + Log (\frac{ A }{ C })\]
i thought A = e^(H-K) x C was correct but it's not...
h - k = log(a/c) log(a/c) = log a - log c so log a - log c = h - k proceed
\[\log(\frac{ A }{ C }) = \log A - \log C \] \[\log A - \log C = h - k\] \[\log A = h - k + \log C\]
ok and then do i use "e" to cancel log?
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@rishavraj
ooo yeahh.... :))
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