\( f(x)=\log _{2}(x+3) \) 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.
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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!