Respuesta :
Given
[tex]\begin{gathered} f(x)=log_2(x+3) \\ and \\ g(x)=log_2(3x+1) \end{gathered}[/tex]a)
[tex]\begin{gathered} f(x)=4 \\ \Rightarrow log_2(x+3)=4 \\ \Leftrightarrow x+3=2^4 \\ \Rightarrow x+3=16 \\ \Rightarrow x=13 \end{gathered}[/tex]The answer to part a) is x=13. The point on the graph is (13,4)
b)
[tex]\begin{gathered} g(x)=4 \\ \Rightarrow log_2(3x+1)=4 \\ \Leftrightarrow3x+1=2^4 \\ \Rightarrow3x+1=16 \\ \Rightarrow3x=15 \\ \Rightarrow x=5 \end{gathered}[/tex]The answer to part b) is x=5, and the point on the graph is (5,4).
c)
[tex]\begin{gathered} f(x)=g(x) \\ \Rightarrow log_2(x+3)=log_2(3x+1) \\ \Rightarrow\frac{ln(x+3)}{ln(2)}=\frac{ln(3x+1)}{ln(2)}] \\ \Rightarrow ln(x+3)=ln(3x+1) \\ \Rightarrow x+3=3x+1 \\ \Rightarrow2x=2 \\ \Rightarrow x=1 \\ and \\ log_2(1+3)=log_2(4)=2 \end{gathered}[/tex]The answer to part c) is x=1 and graphs intersect at (1,2).
d)
[tex]\begin{gathered} (f+g)(x)=7 \\ \Rightarrow log_2(x+3)+log_2(3x+1)=7 \\ \Rightarrow log_2((x+3)(3x+1))=7 \\ \Leftrightarrow(x+3)(3x+1)=2^7 \\ \Rightarrow3x^2+10x+3=128 \\ \Rightarrow3x^2+10x-125=0 \end{gathered}[/tex]Solving the quadratic equation using the quadratic formula,
[tex]\begin{gathered} \Rightarrow x=\frac{-10\pm\sqrt{10^2-4*3*-125}}{3*2} \\ \Rightarrow x=-\frac{25}{3},5 \end{gathered}[/tex]However, notice that if x=-25/3,
[tex]log_2(x+3)=log_2(-\frac{25}{3}+3)=log_2(-\frac{16}{3})\rightarrow\text{ not a real number}[/tex]Therefore, x=-25/3 is not a valid answer.
The answer to part d) is x=5.
e)
[tex]\begin{gathered} log_2(x+3)-log_2(3x+1)=3 \\ log_2(\frac{x+3}{3x+1})=3 \\ \Leftrightarrow\frac{x+3}{3x+1}=2^3=8 \\ \Rightarrow x+3=24x+8 \\ \Rightarrow23x=-5 \\ \Rightarrow x=-\frac{5}{23} \end{gathered}[/tex]The answer to part e) is x=-5/23
