Risolvi log_{10}({81/8})+2log(2/3)log_{10}({3/2})+log_{10}({3/4})

$$\log_{ 10 }({ \frac{ 81 }{ 8 } }) +2 \log ( \frac{ 2 }{ 3 } ) \log_{ 10 }({ \frac{ 3 }{ 2 } }) + \log_{ 10 }({ \frac{ 3 }{ 4 } })$$

Calcola

$$\frac{\ln(\frac{243}{32})\ln(10)-2\ln(\frac{2}{3})^{2}}{\ln(10)^{2}}\approx 0,818440032$$

Scomponi in fattori

$$\frac{\log_{e} {(\frac{243}{32})} \log_{e} {(10)} - 2 \log_{e} {(\frac{2}{3})} ^ {2}}{\log_{e} {(10)} ^ {2}} = 0,8184400322467759$$