Verified Answer :

H=k+log(a/c)
H-k = log(a/c)
e^{h-k}=a/c
a = ce^{h-k}

remember some properties
is no base is stated, assume base 10
and
\(log_a(b)=c\) means \(a^c=b\)
and
\(a^{log_a(b)}=b\)
and
\(log_a(b/c)=log_a(b)-log_a(c)\)
and
\(x^{a+b}=(x^a)(x^b)\)
so

\(H=K+log(a/c)\)
\(H=K+log_{10}(a/c)\)
\(H=K+log_{10}(a)-log_{10}(c)\)
minus K-log10(c) from both sides
\(H-K+log_{10}(c)=log_{10}(a)\)
convert
\(10^{H-K+log_{10}(c)}=a\)
\((10^{H-K}(10^{log_{10}(c)}=a\)
\((10^{H-K})(c)=a\)
\(a=c10^{H-K}\)