Solve for A H=K+ log of A/C

Rewrite the equation as ${\displaystyle K+\log \left(\frac{A}{C}\right)=H}$.

Subtract ${\displaystyle K}$ from both sides of the equation.

Rewrite ${\displaystyle \log \left(\frac{A}{C}\right)=H-K}$ in exponential form using the definition of a logarithm. If ${\displaystyle x}$ and ${\displaystyle b}$ are positive real numbers and ${\displaystyle b\ne 1}$, then ${\displaystyle {\log }_b\left(x\right)=y}$ is equivalent to ${\displaystyle b^y=x}$.