5. Match the exponent equality to the correct solution for \( x \). \( 16^{x+2}=4^{x-2} \) 3. \( 25^{x-5}=5 \) 4. \( 27^{x+3}=5^{x+2} \)

To solve \( 16^{x+2}=4^{x-2} \), we can express both sides as powers of 2: \( 16 = 2^4 \) and \( 4 = 2^2 \). Thus, the equation becomes \( (2^4)^{x+2} = (2^2)^{x-2} \) or \( 2^{4(x+2)} = 2^{2(x-2)} \). This simplifies to \( 4(x+2) = 2(x-2) \). Solving this gives \( x = -3 \). Next, for \( 25^{x-5}=5 \), we rewrite \( 25 \) as \( 5^2 \), resulting in \( (5^2)^{x-5}=5^1 \). Therefore, \( 5^{2(x-5)} = 5^1 \) leads to \( 2(x-5) = 1 \), yielding \( x = 6.5 \) or \( x = \frac{13}{2} \). Finally, we need to examine \( 27^{x+3}=5^{x+2} \) by rewriting \( 27 \) as \( 3^3 \). Hence, \( (3^3)^{x+3}=5^{x+2} \) gives \( 3^{3(x+3)} = 5^{x+2} \), which doesn't yield a product of simple base equality. Therefore, this equation generally requires logs or numerical methods for solutions. Have fun exploring the power of exponents! They pop up everywhere, from simplifying equations to calculating compounds in finance – like determining interest over time, showing just how every little exponent can multiply your returns!