Question

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

Answer 1

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

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