Respuesta :

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
[tex]log_a(b)=c[/tex] means [tex]a^c=b[/tex]
and
[tex]a^{log_a(b)}=b[/tex]
and
[tex]log_a(b/c)=log_a(b)-log_a(c)[/tex]
and
[tex]x^{a+b}=(x^a)(x^b)[/tex]
so

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