Question

f(x) = log 2(x+3) and g(x) = log 2(3x + 1).(a) Solve f(x) = 4. What point is on the graph of f?(b) Solve g(x) = 4. What point is on the graph of g?(c) Solve f(x) = g(x). Do the graphs off and g intersect? If so, where?(d) Solve (f+g)(x) = 7.(e) Solve (f-g)(x) = 3.

Answer 1

Given

`\begin{gathered} f(x)=log_2(x+3) \\ and \\ g(x)=log_2(3x+1) \end{gathered}`

a)

`\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}`

The answer to part a) is x=13. The point on the graph is (13,4)

b)

`\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}`

The answer to part b) is x=5, and the point on the graph is (5,4).

c)

`\begin{gathered} f(x)=g(x) \\ \Rightarrow log_2(x+3)=log_2(3x+1) \\ \Rightarrow(ln(x+3))/(ln(2))=(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}`

The answer to part c) is x=1 and graphs intersect at (1,2).

d)

`\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}`

Solving the quadratic equation using the quadratic formula,

`\begin{gathered} \Rightarrow x=(-10\pm√(10^2-4*3*-125))/(3*2) \\ \Rightarrow x=-(25)/(3),5 \end{gathered}`

However, notice that if x=-25/3,

`log_2(x+3)=log_2(-(25)/(3)+3)=log_2(-(16)/(3))\rightarrow\text{ not a real number}`

Therefore, x=-25/3 is not a valid answer.

The answer to part d) is x=5.

e)

`\begin{gathered} log_2(x+3)-log_2(3x+1)=3 \\ log_2((x+3)/(3x+1))=3 \\ \Leftrightarrow(x+3)/(3x+1)=2^3=8 \\ \Rightarrow x+3=24x+8 \\ \Rightarrow23x=-5 \\ \Rightarrow x=-(5)/(23) \end{gathered}`

The answer to part e) is x=-5/23