Question

$log_{3}( 3^{ 3x^{2} }-16x+18+ \frac{2}{9})= log_{4} 0,25$

Answer 1

$\displaystyle \mathtt{log_3\left(3^{3x^2-16x+18}+ \frac{2}{9} \right)=log_40,25~~~~~~~~~~~~~~~~~~~~~3^{3x^2-16x+18}=t}\\ \\ \mathtt{log_3\left(t+ \frac{2}{9}\right)=log_4 \frac{25}{100}\Rightarrow log_3 \left(t+ \frac{2}{9}\right)=log_4 \frac{1}{4}\Rightarrow }\\ \\ \mathtt{\Rightarrow log_3 \left(t+ \frac{2}{9}\right)=log_44^{-1} \Rightarrow log_3 \left(t+ \frac{2}{9} \right)=-1 \Rightarrow }$
$\displaystyle \mathtt{\Rightarrow log_3 \left(t+ \frac{2}{9}\right)=log_3 \frac{1}{3} \Rightarrow t+ \frac{2}{9} = \frac{1}{3} \Rightarrow 9t+2=3\Rightarrow }\\ \\ \mathtt{\Rightarrow 9t=3-2 \Rightarrow 9t=1 \Rightarrow t= \frac{1}{9} }} }$
$\displaystyle \mathtt{3^{3x^2-16x+18}= \frac{1}{9}} \\ \\ \mathtt{3^{3x^2-16x+18}=3^{-2}}\\ \\ \mathtt{3x^2-16x+18=-2} \\ \\ \mathtt{3x^2-16x+18+2=0} \\ \\ \mathtt{3x^2-16x+20=0}\\ \\ \mathtt{a=3,~b=-16,~c=20}\\ \\ \mathtt{\Delta=b^2-4ac=(-16)^2-4 \cdot 3 \cdot 20=256-240=16\ \textgreater \ 0}$
$\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-(-16)+ \sqrt{16} }{2 \cdot 3}= \frac{16+4}{6}= \frac{20}{6}= \frac{10}{3} } \\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-16)- \sqrt{16} }{2\cdot 3} = \frac{16-4}{6}= \frac{12}{6}=2 }$

Answer 2

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