Question

1. $64 cdot 16^{-3x} = 16^{3x - 2}$

Explanation:

Step1: Rewrite 64 as a power of 16

Since \(64 = 16^{\frac{3}{2}}\) (because \(16^{\frac{3}{2}}=\sqrt{16}^3 = 4^3=64\)), the equation \(64\cdot16^{-3x}=16^{3x - 2}\) can be rewritten as \(16^{\frac{3}{2}}\cdot16^{-3x}=16^{3x - 2}\).

Step2: Use the exponent rule \(a^m\cdot a^n=a^{m + n}\)

For the left - hand side, we have \(16^{\frac{3}{2}-3x}\). So the equation becomes \(16^{\frac{3}{2}-3x}=16^{3x - 2}\).

Step3: Set the exponents equal

Since the bases are the same (\(a^m=a^n\) implies \(m = n\) when \(a>0,a
eq1\)), we set \(\frac{3}{2}-3x=3x - 2\).

Step4: Solve for x

First, add \(3x\) to both sides: \(\frac{3}{2}=6x - 2\).
Then, add 2 to both sides: \(\frac{3}{2}+2=6x\).
\(\frac{3 + 4}{2}=6x\), so \(\frac{7}{2}=6x\).
Finally, divide both sides by 6: \(x=\frac{7}{12}\).

Answer:

\(x = \frac{7}{12}\)