Respuesta :

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[tex]2.3\log(A+300)-5.1=R\ \ \ \ |+5.1\\\\2.3\log(A+300)=R+5.1\ \ \ |:2.3\\\\\log(A+300)=\dfrac{R+5.1}{2.3}\iff A+300=10^{\frac{R+5.1}{2.3}}\ \ \ |-300\\\\\boxed{A=10^{\frac{10R+51}{23}}-300}[/tex]