Question

1. $4^{-2x} \cdot 4^{x} = 64$
2. $6^{-2x} \cdot 6^{-x} = \frac{1}{216}$
3. $2^{x} \cdot \frac{1}{32} = 32$
4. $2^{-3p} \cdot 2^{2p} = 2^{2p}$
5. $64 \cdot 16^{-3x} = 16^{3x - 2}$
6. $\frac{81^{3n + 2}}{243^{-n}} = 3^{4}$
7. $81 \cdot 9^{-2b - 2} = 27$
8. $9^{-3x} \cdot 9^{x} = 27$
9. $\left(\frac{1}{6}\

ight)^{3x + 2} \cdot 216^{3x} = \frac{1}{216}$

1. $243^{k + 2} \cdot 9^{2k - 1} = 9$
2. $16^{r} \cdot 64^{3 - 3r} = 64$
3. $16^{2p - 3} \cdot 4^{-2p} = 2^{4}$

Explanation:

Let's solve problem 13: \( 4^{-2x} \cdot 4^{x} = 64 \)

Step 1: Use exponent rule \( a^m \cdot a^n = a^{m + n} \)

\( 4^{-2x + x} = 64 \)
Simplify the exponent: \( 4^{-x} = 64 \)

Step 2: Express 4 and 64 as powers of 2

We know that \( 4 = 2^2 \) and \( 64 = 2^6 \). So we can rewrite the equation:
\( (2^2)^{-x} = 2^6 \)

Step 3: Use exponent rule \( (a^m)^n = a^{m \cdot n} \)

\( 2^{-2x} = 2^6 \)

Step 4: Since the bases are equal, set the exponents equal

\( -2x = 6 \)

Step 5: Solve for \( x \)

Divide both sides by -2: \( x = \frac{6}{-2} = -3 \)

Answer:

\( x = -3 \)