1. Solve the equation x = (2x+ 3)1/22. Consider f1(x) = ln(x + 1) + ln (x-1) and f2(x) = ln(x^2-1).a) State domains and ranges of f1 and f2.b) Sketch the curves y = f1(x) and y = f2(x).

Answer:1.  [tex]x=3[/tex]; 2. Domain:  [tex]x>1[/tex]  Range: all real numbersStep-by-step explanation:Let's find the solutions. 1. Solve the equation [tex]x=\sqrt{2x+3}[/tex] so: [tex](x)^2=(\sqrt{2x+3})^2[/tex] [tex]x^2=2x+3[/tex] [tex]x^2-2x-3=0[/tex] [tex]x1=\frac{-b+\sqrt{b^{2}-4ac}}{2a}[/tex][tex]x1=\frac{2+\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex][tex]x1=3[/tex][tex]x2=\frac{-b-\sqrt{b^{2}-4ac}}{2a}[/tex][tex]x2=\frac{2-\sqrt{(-2)^{2}-(4*1*(-3))}}{2*1}[/tex][tex]x2=-1[/tex] Although we have two answers, remember that from the original equation the result of [tex]\sqrt{2x+3} > 0[/tex] is never negative. So -1 do not solve the equation.In conlcusion, the equation is solved by x=3.2A. Domains and ranges of f1(x) and f2(x)[tex]f1(x)=ln(x+1)+ln(x-1)[/tex]Using logarithmic property [tex]ln(a)+ln(b)=ln(a*b)[/tex] we have:[tex]f1(x)=ln(x^2-1)[/tex] because:[tex]ln(x)[/tex] is defined by  [tex]x>0[/tex] then:[tex]x^2-1>0[/tex] [tex]x>\sqrt{1}[/tex] so the domain of f1(x) is [tex]x>1[/tex]Now for the range:[tex]f1(x)=ln(x+1)+ln(x-1)[/tex][tex]y=ln(x^2-1)[/tex][tex]e^y=x^2-1[/tex][tex]\sqrt{e^y+1}=x^2-1[/tex] notice that [tex]e^y+1[/tex] is always positive, so the range of f1(x) is all real numbers.Be aware that although point number two of the problem mentioned two equations, f1(x)=f2(x) by logarithmic properties, so their domains and ranges are the same. 2B. Graph of f1(x) is attached. Because f1(x)=f2(x) both functions plot equal.