\( f(x)=\log _{2}(x+3) \) and \( g(x)=\log _{2}(3 x+1) \). (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 of \( f \) and \( g \) intersect? If so, where? (d) Solve \( (f+g)(x)=7 \). (e) Solve \( (f-g)(x)=6 \).

(a) To solve \( f(x) = 4 \): \[ \log_2(x + 3) = 4 \implies x + 3 = 2^4 \implies x + 3 = 16 \implies x = 13. \] So the point on the graph of \( f \) is \( (13, 4) \). (b) To solve \( g(x) = 4 \): \[ \log_2(3x + 1) = 4 \implies 3x + 1 = 2^4 \implies 3x + 1 = 16 \implies 3x = 15 \implies x = 5. \] Thus, the point on the graph of \( g \) is \( (5, 4) \). (c) To solve \( f(x) = g(x) \): \[ \log_2(x + 3) = \log_2(3x + 1) \implies x + 3 = 3x + 1 \implies 2 = 2x \implies x = 1. \] The graphs of \( f \) and \( g \) intersect at the point \( (1, f(1)) = (1, \log_2(4)) = (1, 2) \). (d) To solve \( (f + g)(x) = 7 \): \[ \log_2(x + 3) + \log_2(3x + 1) = 7 \implies \log_2((x + 3)(3x + 1)) = 7. \] This means that: \[ (x + 3)(3x + 1) = 2^7 \implies (x + 3)(3x + 1) = 128. \] Expanding gives \( 3x^2 + 10x + 3 - 128 = 0 \), which simplifies to \( 3x^2 + 10x - 125 = 0 \). Solving this quadratic equation provides the values of \( x \). (e) To solve \( (f - g)(x) = 6 \): \[ \log_2(x + 3) - \log_2(3x + 1) = 6 \implies \log_2\left(\frac{x + 3}{3x + 1}\right) = 6. \] This implies that: \[ \frac{x + 3}{3x + 1} = 2^6 \implies \frac{x + 3}{3x + 1} = 64. \] This leads to \( x + 3 = 64(3x + 1) \), which can be maneuvered to isolate \( x \).