Question

Sa se calculeze $log _{3} 27+log_{4} 16- \sqrt[3]{125}$

Answer 1

ecuatia  devine
log(3)3^3+log(4)4^2-5=
3+2-5+0

Answer 2

$\large\boxed{ \log_a y=x \Leftrightarrow a^x=y} \\ \boxed{\sqrt[3]{a} = b \Leftrightarrow b^3=a} \\\\ log_3 27= x \Leftrightarrow 3^x=27=\ \textgreater \ x=3 \\ \\ log_4 16=x\Leftrightarrow 4^x=16=\ \textgreater \ x=2 \\ \\ \sqrt[3]{125} =b\Leftrightarrow b^3=125 =\ \textgreater \ b=5 \\ \\ 3+2-5=0$