\( f(x)=\log _{2}(x+3) \) 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 \( (f-g)(x)=6 \).

Let's solve each part step by step. ### (a) Solve \( f(x) = 4 \). What point is on the graph of \( f \)? Given: \[ f(x) = \log_2(x + 3) \] Set \( f(x) = 4 \): \[ \log_2(x + 3) = 4 \] Convert the logarithmic equation to its exponential form: \[ x + 3 = 2^4 \] \[ x + 3 = 16 \] \[ x = 16 - 3 \] \[ x = 13 \] **Point on the graph of \( f \):** \[ (13, 4) \] --- ### (b) Solve \( g(x) = 4 \). What point is on the graph of \( g \)? Given: \[ g(x) = \log_2(3x + 1) \] Set \( g(x) = 4 \): \[ \log_2(3x + 1) = 4 \] Convert to exponential form: \[ 3x + 1 = 2^4 \] \[ 3x + 1 = 16 \] \[ 3x = 16 - 1 \] \[ 3x = 15 \] \[ x = \frac{15}{3} \] \[ x = 5 \] **Point on the graph of \( g \):** \[ (5, 4) \] --- ### (c) Solve \( f(x) = g(x) \). Do the graphs of \( f \) and \( g \) intersect? If so, where? Set \( f(x) = g(x) \): \[ \log_2(x + 3) = \log_2(3x + 1) \] Since the logarithms are equal and have the same base, their arguments must be equal: \[ x + 3 = 3x + 1 \] Solve for \( x \): \[ x + 3 = 3x + 1 \] \[ 3 - 1 = 3x - x \] \[ 2 = 2x \] \[ x = 1 \] **Verify the solution is within the domain:** \[ x + 3 > 0 \Rightarrow 1 + 3 = 4 > 0 \] \[ 3x + 1 > 0 \Rightarrow 3(1) + 1 = 4 > 0 \] Both conditions are satisfied. **Find the corresponding \( y \):** \[ f(1) = \log_2(1 + 3) = \log_2(4) = 2 \] **Intersection Point:** \[ (1, 2) \] **Conclusion:** Yes, the graphs of \( f \) and \( g \) intersect at the point \( (1, 2) \). --- ### (d) Solve \( (f + g)(x) = 7 \). Given: \[ (f + g)(x) = f(x) + g(x) = \log_2(x + 3) + \log_2(3x + 1) \] Set equal to 7: \[ \log_2(x + 3) + \log_2(3x + 1) = 7 \] Combine the logarithms: \[ \log_2\left( (x + 3)(3x + 1) \right) = 7 \] Convert to exponential form: \[ (x + 3)(3x + 1) = 2^7 \] \[ (x + 3)(3x + 1) = 128 \] Expand the left side: \[ 3x^2 + x + 9x + 3 = 128 \] \[ 3x^2 + 10x + 3 = 128 \] Bring all terms to one side: \[ 3x^2 + 10x - 125 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-125)}}{2 \cdot 3} \] \[ x = \frac{-10 \pm \sqrt{100 + 1500}}{6} \] \[ x = \frac{-10 \pm \sqrt{1600}}{6} \] \[ x = \frac{-10 \pm 40}{6} \] **Solutions:** \[ x = \frac{30}{6} = 5 \] \[ x = \frac{-50}{6} = -\frac{25}{3} \approx -8.333 \] **Check the domain:** \[ x = 5 \quad \text{is valid since} \quad 5 > -\frac{1}{3} \] \[ x = -\frac{25}{3} \quad \text{is invalid since} \quad -\frac{25}{3} < -\frac{1}{3} \] **Solution:** \[ x = 5 \] --- ### (e) Solve \( (f - g)(x) = 6 \). Given: \[ (f - g)(x) = f(x) - g(x) = \log_2(x + 3) - \log_2(3x + 1) \] Set equal to 6: \[ \log_2(x + 3) - \log_2(3x + 1) = 6 \] Combine the logarithms: \[ \log_2\left( \frac{x + 3}{3x + 1} \right) = 6 \] Convert to exponential form: \[ \frac{x + 3}{3x + 1} = 2^6 \] \[ \frac{x + 3}{3x + 1} = 64 \] Solve for \( x \): \[ x + 3 = 64(3x + 1) \] \[ x + 3 = 192x + 64 \] Bring all terms to one side: \[ x - 192x + 3 - 64 = 0 \] \[ -191x - 61 = 0 \] \[ 191x = -61 \] \[ x = -\frac{61}{191} \] **Simplify:** \[ x = -\frac{61}{191} \approx -0.319 \] **Check the domain:** \[ x + 3 > 0 \Rightarrow -\frac{61}{191} + 3 > 0 \quad \text{(True)} \] \[ 3x + 1 > 0 \Rightarrow 3\left(-\frac{61}{191}\right) + 1 > 0 \quad \text{(True)} \] **Solution:** \[ x = -\frac{61}{191} \] --- ### Summary of Solutions - **(a)** \( x = 13 \). Point: \( (13, 4) \) - **(b)** \( x = 5 \). Point: \( (5, 4) \) - **(c)** The graphs intersect at \( (1, 2) \) - **(d)** \( x = 5 \) - **(e)** \( x = -\dfrac{61}{191} \)