[tex]\displaystyle\bf\\a)\\log_3(5x-7)=log_3(2x-1)\\5x-7=2x-1\\5x-2x=-1+7\\3x=6\\x=6:3=2\\\\b)\\log_2(x^2-5x+9)=log_2(x+4)\\x^2-5x+9=x+4\\x^2-5x+9-x-4=0\\x^2-6x+5=0\\x^2-x-5x+5=0\\x(x-1)-5(x-1)=0\\(x-1)(x-5)=0\\x-1=0~\implies x_1=1\\x-5=0~\implies x_2=5\\\\c)\\log_5(x^2+9)=2\\x^2+9=5^2\\x^2+9=25\\x^2=25-9\\x^2=16\\x_{12}=\pm\sqrt{16}\\x_1=-4\\x_2=4[/tex]
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[tex]\displaystyle\bf\\d)\\lg^2(x+1)-4lg(x+1)+3=0\\Substitutie\!:~~lg(x+1)=y\\y^2-4y+3=0\\y^2-y-3y+3=0\\y(y-1)-3(y-1)=0\\(y-1)(y-3)=0\\y-1=0~\implies~y_1=1\\y-3=0~\implies~y_2=3\\Ne~intoarcem~la~substitutie.\\lg(x+1)=y\\lg(x+1)=1~\implies~x+1=10^1~\implies~x=10-1=9\\lg(x+1)=3~\implies~x+1=10^3~\implies~x=1000-1=999[/tex]