Question

Решить lg2+lg3/lg27+lg12

Answer 1

$\frac{lg\, 2+lg\, 3}{lg27+lg12}=\frac{lg\, 6}{lg324}=log\, _{324}\, 6=\frac{1}{log_6324}=\frac{1}{log_6(6^2\cdot 9)}=\frac{1}{log_66^2+log_63^2}=\\\\=\frac{1}{2+2log_63}=\frac{1}{2\, (1+log_63)}=\frac{1}{2\, (1+\frac{1}{log_36})}=\frac{1}{2\cdot (1+\frac{1}{1+log_32})}=\frac{1}{2\cdot \frac{2+log_32}{1+log_32}}=\\\\=\frac{1+log_32}{2\cdot (2+log_32)}$

$lg2+\frac{lg3}{lg27}+lg12=lg2+\frac{lg3}{lg3^3}+lg(3\cdot 2^2)=\\\\=lg2+\frac{lg3}{3lg3}+lg3+2lg2=3\, lg2+lg3+\frac{1}{3}$

Answer 2

Объяснение: lg2 + (lg3/lg27)+lg12=lg(2×12)+(lg3/3lg3) = lg24 + 1/3.