Solve each of the following inequalities for \( x \), where \( x \) is an integer. \( \begin{array}{ll}\text { a. } 2^{x} \geq 8 & \text { c. } 5^{4 x} \leq 3125 \\ \text { b. } 81^{x} \geq 27 & \text { d. } 6^{x}<1\end{array} \)
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To solve each inequality for \( x \): a. \( 2^{x} \geq 8 \) Since \( 8 = 2^3 \), we rewrite the inequality as \( 2^{x} \geq 2^3 \). This implies \( x \geq 3 \). As \( x \) is an integer, the solutions are \( x = 3, 4, 5, \ldots \). b. \( 81^{x} \geq 27 \) We can express both numbers as powers of 3: \( 81 = 3^4 \) and \( 27 = 3^3 \). Thus, \( (3^4)^{x} \geq 3^3 \) simplifies to \( 3^{4x} \geq 3^3 \). This leads to \( 4x \geq 3 \), so \( x \geq \frac{3}{4} \). The integer solutions are \( x = 1, 2, 3, \ldots \). c. \( 5^{4x} \leq 3125 \) Since \( 3125 = 5^5 \), we can rewrite the inequality as \( 5^{4x} \leq 5^5 \), which implies \( 4x \leq 5 \). Solving this gives \( x \leq \frac{5}{4} \). The integer solutions are \( x = 0, 1 \). d. \( 6^{x} < 1 \) The inequality \( 6^{x} < 1 \) is only satisfied when \( x < 0 \) (since \( 6^0 = 1 \)). Therefore, the integer solutions are \( x = -1, -2, -3, \ldots \).
