Question

$log_{3} ( 3^{ x^{2-13x+28}}+ \frac{2}{9})= log_{5} 0,2$

Answer 1

㏒₅1/5=㏒₅5⁻¹=-㏒₅5=-1
3ˣ²⁻¹³ˣ⁺²⁸+2/9=3⁻¹
3ˣ²⁻¹³ˣ⁺²⁸+2/9=1/3
9·3ˣ²-¹³ˣ⁺²⁸=1
3ˣ²⁻¹³ˣ⁺³⁰=3⁰
x²-13x+30=0
Δ=49
√Δ=7
x₁=10
x₂=3

Answer 2

$\displaystyle \mathtt{log_3 \left(3^{x^2-13x+28}+ \frac{2}{9} \right)=log_50,2~~~~~~~~~~~~~~~~~~~~~3^{x^2-13x+28}=t} \\ \\ \mathtt{log_3 \left(t+ \frac{2}{9}\right) =log_5 \frac{2}{10}}\Rightarrow \mathtt{log_3 \left(t+ \frac{2}{9}\right)=log_5 \frac{1}{5}\Rightarrow } \\ \\ \mathtt{\Rightarrow log_3\left(t+ \frac{2}{9}\right)=log_55^{-1}\Rightarrow log_3 \left(t+ \frac{2}{9}\right)=-1\Rightarrow }$
$\displaystyle \mathtt{\Rightarrow log_3 \left(t+ \frac{2}{9}\right)=log_3 \frac{1}{3}\Rightarrow t+ \frac{2}{9}= \frac{1}{3} \Rightarrow 9t+2=3 \Rightarrow } \\ \\ \mathtt{\Rightarrow 9t=3-2 \Rightarrow 9t=1 \Rightarrow t= \frac{1}{9} } \\ \\ \mathtt{3^{x^2-13x+28}= \frac{1}{9} }\\ \\ \mathtt{3^{x^2-13x+28}=3^{-2}}\\ \\ \mathtt{x^2-13x+28=-2} \\ \\ \mathtt{x^2-13x+28+2=0}\\ \\ \mathtt{x^2-13x+30=0}\\ \\ \mathtt{a=1,~b=-13,~c=30}\\ \\ \mathtt{\Delta=b^2-4ac=(-13)^2-4 \cdot 1 \cdot 30=169-120=49\ \textgreater \ 0}$

$\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-13)+ \sqrt{49} }{2 \cdot 1} = \frac{13+7}{2}= \frac{20}{2}=10 }\\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-13)- \sqrt{49} }{2 \cdot 1}= \frac{13-7}{2}= \frac{6}{2} =3 }$