\( f(x)=\log _{2}(x+3) \) 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 \( (f-g)(x)=5 \).

Let's tackle these logarithmic equations step-by-step! (a) To solve \( f(x) = 4 \), we set up the equation: \[ \log_2(x + 3) = 4 \] Converting from logarithmic to exponential form, we get: \[ x + 3 = 2^4 \implies x + 3 = 16 \implies x = 13 \] Thus, the point on the graph of \( f \) is \( (13, 4) \). (b) Now let's solve \( g(x) = 4 \): \[ \log_2(3x + 1) = 4 \] Again, converting to exponential form: \[ 3x + 1 = 2^4 \implies 3x + 1 = 16 \implies 3x = 15 \implies x = 5 \] So, the point on the graph of \( g \) is \( (5, 4) \). (c) To find where \( f(x) = g(x) \): \[ \log_2(x + 3) = \log_2(3x + 1) \] Since the logarithmic functions equal each other, their insides must match: \[ x + 3 = 3x + 1 \] Rearranging gives: \[ 2 = 2x \implies x = 1 \] Now, substituting back to find \( y \): \[ f(1) = \log_2(4) = 2 \quad \text{and} \quad g(1) = \log_2(4) = 2. \] Thus, the graphs of \( f \) and \( g \) intersect at the point \( (1, 2) \). (d) Next, we solve \( (f+g)(x) = 7 \): \[ f(x) + g(x) = \log_2(x + 3) + \log_2(3x + 1) = \log_2((x + 3)(3x + 1)) = 7. \] This implies: \[ (x + 3)(3x + 1) = 2^7 = 128. \] Expanding and simplifying: \[ 3x^2 + 10x + 3 = 128 \implies 3x^2 + 10x - 125 = 0. \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-125)}}{2 \cdot 3} = \frac{-10 \pm \sqrt{100 + 1500}}{6} = \frac{-10 \pm \sqrt{1600}}{6} = \frac{-10 \pm 40}{6}. \] This gives us two solutions: \( x = 5 \) and \( x = -\frac{25}{3} \) (which we discard since the log function is undefined for negative inputs). Therefore, we have \( x = 5 \). (e) Finally, we solve \( (f-g)(x) = 5 \): \[ f(x) - g(x) = \log_2(x + 3) - \log_2(3x + 1) = \log_2\left(\frac{x + 3}{3x + 1}\right) = 5. \] Then we convert to exponential form: \[ \frac{x + 3}{3x + 1} = 2^5 = 32. \] Cross multiplying gives: \[ x + 3 = 32(3x + 1) \implies x + 3 = 96x + 32 \implies 3 - 32 = 95x \implies 95x = -29 \implies x = -\frac{29}{95}. \] This is a valid solution within the domain of the logarithmic functions for input \( x + 3 > 0 \) and \( 3x + 1 > 0 \).