EXERCISE 6 Draw the graphs of \( f(x) \) and \( g(x) \) on the same set of axes and use them to solve the inequality. 1 \( f(x)=\log _{3} x \) and \( g(x)=2 \). Now solve \( \log _{3} x \leq 2 \) \( 2 f(x)=\log _{\frac{1}{4}} x \) and \( g(x)=-1 \). Now solve \( \log _{\frac{1}{4}} x \geq-1 \) \( 3 f(x)=\log _{5} x \) and \( g(x)=1 \). Now solve \( \log _{5} x \geq 1 \) \( 4 f(x)=\log _{\frac{1}{2}} x \) and \( g(x)=x-1 \). Now solve \( \log _{\frac{1}{2}} x \geq x-1 \) \( f(x)=\log _{4} x \) and \( g(x)=-2 x+2 \). Now solve \( \log _{4} x \leq-2 x+2 \)
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Mind Expander
To tackle these inequalities, let's take a look at the graphs. For \( f(x) = \log_{3} x \) and \( g(x) = 2 \), you will find that \( f(x) \) crosses the line \( y = 2 \) at \( x = 9 \). Thus, the solution to \( \log_{3} x \leq 2 \) is \( x \in (0, 9] \). Next, when solving \( \log_{\frac{1}{4}} x \geq -1 \) using \( f(x) = \log_{\frac{1}{4}} x \) and \( g(x) = -1 \), the critical point occurs when \( x = 4 \) since that's where the graph of \( f(x) \) meets \( g(x) \). The solution is \( x \in [4, \infty) \). For \( \log_{5} x \geq 1 \), the crossing point happens at \( x = 5 \). Hence, the solution is \( x \in [5, \infty) \) when you draw the graphs accordingly. Moving to \( \log_{\frac{1}{2}} x \geq x - 1 \), this involves analyzing the concavity of the two functions. The solution will be around the intersection points obtained graphically, giving you a range depending on where these graphs cross. Finally, looking at \( \log_{4} x \leq -2x + 2 \), the intersection also provides significant bounds for the solution set. The joy of graphing these functions lies in the clarity they bring to the inequalities! Happy graphing!
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