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\frac{\log_{10}\left(81\right)-6\log_{10}\left(2\right)-2\log_{10}\left(z\right)}{2}
Factor out \frac{1}{2}.
\frac{-2\ln(z)+\ln(\frac{81}{64})}{\ln(10)}
Consider \ln(81)\times \frac{1}{\ln(10)}-6\ln(2)\times \frac{1}{\ln(10)}-2\ln(z)\times \frac{1}{\ln(10)}. Factor out \frac{1}{\ln(10)}.
\frac{-2\ln(z)+\ln(\frac{81}{64})}{2\ln(10)}
Rewrite the complete factored expression.