\( f(x)=\log _{2}(x+3) \) 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:** \[ f(x) = \log_{2}(x + 3) \] \[ f(x) = 4 \] **Solution:** 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:** \[ (13, 4) \] --- ### **(b) Solve \( g(x) = 4 \). What point is on the graph of \( g \)?** **Given:** \[ g(x) = \log_{2}(3x + 1) \] \[ g(x) = 4 \] **Solution:** 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:** \[ (5, 4) \] --- ### **(c) Solve \( f(x) = g(x) \). Do the graphs of \( f \) and \( g \) intersect? If so, where?** **Given:** \[ f(x) = \log_{2}(x + 3) \] \[ g(x) = \log_{2}(3x + 1) \] **Solution:** Set \( f(x) = g(x) \): \[ \log_{2}(x + 3) = \log_{2}(3x + 1) \] Since the logarithmic functions are equal, their arguments must be equal (assuming they are within the domain): \[ x + 3 = 3x + 1 \] \[ x - 3x = 1 - 3 \] \[ -2x = -2 \] \[ x = 1 \] Now, find \( y \) using \( f(x) \) or \( g(x) \): \[ f(1) = \log_{2}(1 + 3) = \log_{2}(4) = 2 \] **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:** \[ (f + g)(x) = f(x) + g(x) = \log_{2}(x + 3) + \log_{2}(3x + 1) = 7 \] **Solution:** Combine the logarithms using the product rule: \[ \log_{2}\left( (x + 3)(3x + 1) \right) = 7 \] Convert to exponential form: \[ (x + 3)(3x + 1) = 2^7 \] \[ (x + 3)(3x + 1) = 128 \] Expand the left side: \[ 3x^2 + x + 9x + 3 = 128 \] \[ 3x^2 + 10x + 3 = 128 \] Subtract 128 from both sides: \[ 3x^2 + 10x - 125 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-125)}}{2 \cdot 3} \] \[ x = \frac{-10 \pm \sqrt{100 + 1500}}{6} \] \[ x = \frac{-10 \pm \sqrt{1600}}{6} \] \[ x = \frac{-10 \pm 40}{6} \] This gives two solutions: \[ x = \frac{30}{6} = 5 \quad \text{and} \quad x = \frac{-50}{6} = -\frac{25}{3} \approx -8.333 \] **Domain Consideration:** - For \( f(x) = \log_{2}(x + 3) \), \( x + 3 > 0 \) ⇒ \( x > -3 \). - For \( g(x) = \log_{2}(3x + 1) \), \( 3x + 1 > 0 \) ⇒ \( x > -\frac{1}{3} \). **Valid Solution:** \[ x = 5 \] (The other solution \( x = -\frac{25}{3} \) is less than \( -\frac{1}{3} \) and thus not in the domain.) --- ### **(e) Solve \( (f - g)(x) = 6 \).** **Given:** \[ (f - g)(x) = f(x) - g(x) = \log_{2}(x + 3) - \log_{2}(3x + 1) = 6 \] **Solution:** Use the logarithmic quotient rule: \[ \log_{2}\left( \frac{x + 3}{3x + 1} \right) = 6 \] Convert to exponential form: \[ \frac{x + 3}{3x + 1} = 2^6 \] \[ \frac{x + 3}{3x + 1} = 64 \] Multiply both sides by \( 3x + 1 \): \[ x + 3 = 64(3x + 1) \] \[ x + 3 = 192x + 64 \] Bring like terms together: \[ x - 192x = 64 - 3 \] \[ -191x = 61 \] \[ x = -\frac{61}{191} \] \[ x \approx -0.319 \] **Domain Consideration:** - For \( f(x) = \log_{2}(x + 3) \), \( x > -3 \). - For \( g(x) = \log_{2}(3x + 1) \), \( x > -\frac{1}{3} \). Check \( x = -\frac{61}{191} \) (approximately -0.319): \[ -0.319 > -0.333 \quad (\text{which is } -\frac{1}{3}) \] Thus, it is within the domain. **Solution:** \[ x = -\frac{61}{191} \approx -0.319 \] --- ### **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)** \( x = -\frac{61}{191} \approx -0.319 \)