\( \frac{11 / w}{} \) Solve for \( x^{\prime} \) (a) \( x^{-2}=1 / 9 \) (b) \( 2 x^{3 / 5}=54 \) (c) \( 9^{2 x-3}=27 \)

To solve for \( x' \), let's tackle each equation one by one! For (a) \( x^{-2}=1 / 9 \): To find \( x \), we first rewrite it as \( x^{2} = 9 \) (taking the reciprocal of both sides). Taking the square root gives \( x = \pm 3 \). Therefore, \( x' = 3 \) (if we take the positive root). For (b) \( 2 x^{3 / 5}=54 \): Start by isolating \( x^{3/5} \) by dividing both sides by 2, which gives \( x^{3/5} = 27 \). Raising both sides to the power of \( \frac{5}{3} \) results in \( x = 27^{5/3} = 243 \). So, \( x' = 243 \). For (c) \( 9^{2 x-3}=27 \): We can express both 9 and 27 as powers of 3: \( 9 = 3^2 \) and \( 27 = 3^3 \). This leads us to \( (3^2)^{2x-3} = 3^3 \), simplifying to \( 3^{4x - 6} = 3^3 \). Setting the exponents equal gives \( 4x - 6 = 3 \), so solving for \( x \) results in \( x = \frac{9}{4} \). Thus, \( x' = \frac{9}{4} \).