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\( f(x)=\log _{2}(x+3) \) and \( g(x)=\log _{2}(3 x+1) \). (a) Solve \( f(x)=4 \). What point is on the graph of \( t \) ? (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 \( (f-g)(x)=5 \). (a) The solution set for \( f(x)=4 \) is \{ \( \} \). (Type an integer or a simplified fraction.) The point where \( f(x)=4 \) is \( \square \). (Type an ordered pair.) (b) The solution set for \( g(x)=4 \) is \{ \}. (Type an integer or a simplified fraction.) The point where \( g(x)=4 \) is \( \square \). (Type an ordered pair.) (c) The solution set for \( f(x)=g(x) \) is \( \{\square\} \). (Type an integer or a simplified fraction.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. Yes, the graphs do intersect at \( \square \). (Type an ordered pair.) B. No, the graphs do not intersect each other.

Ask by Murray Lindsey. in the United States
Jan 26,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

**(a)** - **Solution Set:** \(\{13\}\) - **Point:** \((13, 4)\) **(b)** - **Solution Set:** \(\{5\}\) - **Point:** \((5, 4)\) **(c)** - **Solution Set:** \(\{1\}\) - **Conclusion:** **A.** Yes, the graphs intersect at \((1, 2)\).

Solution

Let's solve each part step by step. --- ### **(a) Solve \( f(x) = 4 \) and Identify the Point on the Graph of \( f \)** **Given:** \[ f(x) = \log_{2}(x + 3) \] **Equation to Solve:** \[ \log_{2}(x + 3) = 4 \] **Solution:** 1. Rewrite the logarithmic equation in exponential form: \[ x + 3 = 2^4 \] 2. Calculate \( 2^4 \): \[ x + 3 = 16 \] 3. Solve for \( x \): \[ x = 16 - 3 \] \[ x = 13 \] **Solution Set:** \[ \{13\} \] **Point on the Graph:** \[ (13, 4) \] --- ### **(b) Solve \( g(x) = 4 \) and Identify the Point on the Graph of \( g \)** **Given:** \[ g(x) = \log_{2}(3x + 1) \] **Equation to Solve:** \[ \log_{2}(3x + 1) = 4 \] **Solution:** 1. Rewrite the logarithmic equation in exponential form: \[ 3x + 1 = 2^4 \] 2. Calculate \( 2^4 \): \[ 3x + 1 = 16 \] 3. Solve for \( x \): \[ 3x = 16 - 1 \] \[ 3x = 15 \] \[ x = 5 \] **Solution Set:** \[ \{5\} \] **Point on the Graph:** \[ (5, 4) \] --- ### **(c) Solve \( f(x) = g(x) \) and Determine if the Graphs Intersect** **Equations:** \[ f(x) = \log_{2}(x + 3) \] \[ g(x) = \log_{2}(3x + 1) \] **Set the Equations Equal:** \[ \log_{2}(x + 3) = \log_{2}(3x + 1) \] **Solution:** 1. Since the logarithmic functions are equal, their arguments must be equal: \[ x + 3 = 3x + 1 \] 2. Solve for \( x \): \[ x + 3 = 3x + 1 \] \[ 3 - 1 = 3x - x \] \[ 2 = 2x \] \[ x = 1 \] 3. Find the corresponding \( y \)-value using \( f(x) \) or \( g(x) \): \[ f(1) = \log_{2}(1 + 3) = \log_{2}(4) = 2 \] \[ g(1) = \log_{2}(3(1) + 1) = \log_{2}(4) = 2 \] **Solution Set:** \[ \{1\} \] **Do the Graphs Intersect?** **Answer:** **A.** Yes, the graphs do intersect at \( (1, 2) \). --- **Summary of Answers:** - **(a)** - **Solution Set:** \(\{13\}\) - **Point:** \((13, \ 4)\) - **(b)** - **Solution Set:** \(\{5\}\) - **Point:** \((5, \ 4)\) - **(c)** - **Solution Set:** \(\{1\}\) - **Conclusion:** **A.** Yes, the graphs do intersect at \((1, \ 2)\).

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Bonus Knowledge

To solve \( f(x) = 4 \), we set up the equation \( \log_2(x + 3) = 4 \). Rewriting this in exponential form, we have \( x + 3 = 2^4 \), which simplifies to \( x + 3 = 16 \). Therefore, \( x = 16 - 3 = 13 \). The point on the graph of \( f \) is \( (13, 4) \). To solve \( g(x) = 4 \), we use a similar approach: \( \log_2(3x + 1) = 4 \) leads to \( 3x + 1 = 16 \). Solving this gives \( 3x = 15 \) or \( x = 5 \). The point on the graph of \( g \) is \( (5, 4) \). For \( f(x) = g(x) \), we set up the equation \( \log_2(x + 3) = \log_2(3x + 1) \). Since the logs are equal, we can equate their arguments: \( x + 3 = 3x + 1 \). Solving this gives \( 2 = 2x \) or \( x = 1 \). The solution set is \( \{1\} \). Yes, the graphs intersect at the point \( (1, \log_2(4)) = (1, 2) \). Next, for \( (f + g)(x) = 7 \), we need to find \( f(x) + g(x) = \log_2(x + 3) + \log_2(3x + 1) = 7 \). Combining the logs leads to \( \log_2((x + 3)(3x + 1)) = 7 \), giving us \( (x + 3)(3x + 1) = 2^7 = 128 \). Expanding and simplifying this yields a quadratic equation that can be solved for \( x \). Lastly, for \( (f - g)(x) = 5 \), we have \( \log_2(x + 3) - \log_2(3x + 1) = 5 \). This can also be rewritten as \( \log_2\left(\frac{x + 3}{3x + 1}\right) = 5 \), leading to \( \frac{x + 3}{3x + 1} = 2^5 = 32 \). Cross-multiplying and simplifying gives another solvable equation. That's a whirlwind tour of logarithmic equations, turning them into quadratic adventures! Just remember: logarithmic functions can hide their secrets behind their logarithmic masks, but with a little algebraic magic, the truth comes out!

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