21. \( h(x)=\left(\frac{2}{3}\right)^{2 x-1} ; h(1.5), h(-2.5) \)

To find \( h(1.5) \) and \( h(-2.5) \), we'll plug these values into the function \( h(x) = \left(\frac{2}{3}\right)^{2x-1} \). Starting with \( h(1.5) \): \[ h(1.5) = \left(\frac{2}{3}\right)^{2(1.5)-1} = \left(\frac{2}{3}\right)^{3-1} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9} \] Next, for \( h(-2.5) \): \[ h(-2.5) = \left(\frac{2}{3}\right)^{2(-2.5)-1} = \left(\frac{2}{3}\right)^{-5-1} = \left(\frac{2}{3}\right)^{-6} = \left(\frac{3}{2}\right)^{6} = \frac{729}{64} \] Thus, we have: - \( h(1.5) = \frac{4}{9} \) - \( h(-2.5) = \frac{729}{64} \)