\( 1 \leftarrow \quad \) Solve each of the following inequalities for \( x \), where \( x \) is an integer. \( \begin{array}{ll}\text { a. } 2^{x} \leq 64 & \text { c. } 3^{4 x}>243 \\ \text { b. } 8^{x}<4 & \text { d. } 2^{x}>1\end{array} \) a. \( x \leq 6 \) (Type an inequality. Use integers for any numbers in the expression.) b. \( x \leq 0 \) (Type an inequality. Use integers for any numbers in the expression.) c. \( \square \) (Type an inequality. Use integers for any numbers in the expression.)
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To solve the inequalities, let's break them down: a. For \( 2^{x} \leq 64 \): - Since \( 64 = 2^{6} \), we can rewrite the inequality as \( 2^{x} \leq 2^{6} \). - Therefore, \( x \leq 6 \). b. For \( 8^{x} < 4 \): - Rewrite \( 8 \) as \( 2^{3} \) and \( 4 \) as \( 2^{2} \), yielding \( (2^{3})^{x} < 2^{2} \) or \( 2^{3x} < 2^{2} \). - Thus, \( 3x < 2 \) leads to \( x < \frac{2}{3} \). Since \( x \) must be an integer, \( x \leq 0 \). c. For \( 3^{4x} > 243 \): - Recognizing that \( 243 = 3^{5} \), we can rewrite it as \( 3^{4x} > 3^{5} \). - This simplifies to \( 4x > 5 \), resulting in \( x > \frac{5}{4} \) or \( x > 1.25 \). As \( x \) is an integer, \( x \geq 2 \). Thus, the answers are: a. \( x \leq 6 \) b. \( x \leq 0 \) c. \( x \geq 2 \)