23) \( \left(\frac{1}{16}\right)^{3 n-3} \cdot 16^{3 n-2}=64^{-n-3} \)

To solve the equation \( \left(\frac{1}{16}\right)^{3 n-3} \cdot 16^{3 n-2}=64^{-n-3} \), we start by expressing everything in terms of powers of 2. First, note that: \[ 16 = 2^4 \quad \text{and thus} \quad \frac{1}{16} = 2^{-4} \] We can rewrite the left-hand side: \[ \left(\frac{1}{16}\right)^{3 n-3} = (2^{-4})^{3 n-3} = 2^{-4(3n-3)} = 2^{-12n + 12} \] \[ 16^{3 n-2} = (2^4)^{3 n-2} = 2^{4(3n-2)} = 2^{12n - 8} \] Now, multiplying these together: \[ \left(\frac{1}{16}\right)^{3 n-3} \cdot 16^{3 n-2} = 2^{-12n + 12} \cdot 2^{12n - 8} = 2^{-12n + 12 + 12n - 8} = 2^{4} \] Now, for the right-hand side: \[ 64 = 2^6 \quad \text{so} \quad 64^{-n-3} = (2^6)^{-n-3} = 2^{-6(n+3)} = 2^{-6n - 18} \] Now, we set the two sides equal: \[ 2^{4} = 2^{-6n - 18} \] By equating the exponents: \[ 4 = -6n - 18 \] Next, we solve for \( n \): \[ 6n = -18 - 4 \] \[ 6n = -22 \] \[ n = -\frac{22}{6} = -\frac{11}{3} \] Thus, the solution to the equation is: \[ \boxed{-\frac{11}{3}} \]