Question

Найдите корень уравнения 2^log3(9x+9)=6

Answer 1

$2^{log_3(9x+9)}=6\; ,\qquad ODZ:\; 9x+9>0\; ,\; \; >-1\\\\2^{log_3(9x+9)}=2^{log_26}\; \; \; \; \; \Big [\; a^{log_{a}b}=b\; ,\; \; a>0\; ,\; a\ne 1\; ,\; b>0\; \Big ]\\\\log_3(9x+9)=log_26\; \; \; \; \Big [\; log_{a}x=b\; \; \to \; \; x=a^{b}\; \Big ]\\\\9x+9=3^{log_26}\\\\9x=3^{log_26}-9\\\\x=\frac{1}{9}\cdot 3^{log_26}-1\\\\x=\frac{3^{log_26}}{3^2}-1\\\\\underline {x=3^{log_26-2}-1}\\\\x=3^{log_2(2\cdot 3)-2}-1\\\\x=3^{log_22-log_23-2}-1\\\\x=3^{1-log_23-2}-1\\\\\underline {x=3^{-log_23-1}-1}$

$P.S.\; \; x=3^{log_26-2}-1\\\\x=3^{\frac{log_36}{log_32}}\cdot 3^{-2}-1\\\\x=\Big (3^{log_36}\Big )^{\frac{1}{log_32}}\cdot \frac{1}{9}-1\\\\x=6^{log_23}\cdot \frac{1}{9}-1\\\\x=\frac{6^{log_23}}{9}-1$