Найдите значение выражения 3^{2+\log_916 }32+log916
аlogab=bа^{\log_ab}=bаlogab=b
logacb=1clogab=logab1c{\log_{a^c}b=\displaystyle\frac{1}{c}\log_ab= \log_ab^{\displaystyle\frac{1}{c}}}logacb=c1logab=logabc1
32+log916=(32)⋅(3log916)=9⋅30,5log316=9⋅3log316=9⋅4=363^{2+\log_916 }= (3^2)\cdot(3^{\log_916})=9\cdot 3^{0,5\log_316}= 9\cdot 3^{\log_3\sqrt{16}}=9\cdot4=3632+log916=(32)⋅(3log916)=9⋅30,5log316=9⋅3log316=9⋅4=36