I7 \( b^{x}=5 \) calculate \( \cdot \frac{18}{2^{-x}} \)

To solve for the expression \( \frac{18}{2^{-x}} \), first, recognize that \( 2^{-x} = \frac{1}{2^x} \), so we can rewrite the expression as \( 18 \cdot 2^x \). From the equation \( b^{x} = 5 \), you want to express \( x \) in terms of \( b \) or vice versa, but it seems we don't have a specific value for \( b \). Assuming \( b \) is something like 2, we can derive \( 2^x = 5 \). Thus our expression becomes: \[ 18 \cdot 2^x = 18 \cdot 5 = 90. \] So, \( \frac{18}{2^{-x}} = 90 \) when \( b = 2 \). If \( b \) is different, the final answer will vary accordingly!