Question

1. $16^{2p - 3} cdot 4^{-2p} = 2^4$

Explanation:

Step1: Express bases as powers of 2

Since \(16 = 2^4\) and \(4 = 2^2\), rewrite the equation:
\((2^4)^{2p - 3} \cdot (2^2)^{-2p} = 2^4\)

Step2: Apply exponent rules (\((a^m)^n = a^{mn}\))

Simplify each term:
\(2^{4(2p - 3)} \cdot 2^{2(-2p)} = 2^4\)
\(2^{8p - 12} \cdot 2^{-4p} = 2^4\)

Step3: Apply exponent rule (\(a^m \cdot a^n = a^{m + n}\))

Combine the exponents on the left:
\(2^{(8p - 12) + (-4p)} = 2^4\)
\(2^{4p - 12} = 2^4\)

Step4: Set exponents equal (since bases are equal)

If \(a^m = a^n\), then \(m = n\):
\(4p - 12 = 4\)

Step5: Solve for \(p\)

Add 12 to both sides:
\(4p = 4 + 12\)
\(4p = 16\)
Divide by 4:
\(p = \frac{16}{4}\)
\(p = 4\)

Answer:

\(p = 4\)