Question

if \\(\frac{1}{64} = 4^{2s-1} \cdot 16^{2s+2}\\), what is the value of \\(s\\)?

-1
0
1
no solution

Explanation:

Express all terms with base 4

We rewrite each part of the equation using base 4.
Given equation:

$$ \frac{1}{64} = 4^{2s-1} \cdot 16^{2s+2} $$

We know that:

$$ 64 = 4^3 \implies \frac{1}{64} = 4^{-3} $$

$$ 16 = 4^2 \implies 16^{2s+2} = (4^2)^{2s+2} $$

Simplify the exponents

We apply exponent rules to simplify the right side.
Using the power of a power rule:

$$ (4^2)^{2s+2} = 4^{2(2s+2)} = 4^{4s+4} $$

Now, substitute these back into the equation:

$$ 4^{-3} = 4^{2s-1} \cdot 4^{4s+4} $$

Using the product rule for exponents:

$$ 4^{-3} = 4^{(2s-1) + (4s+4)} $$

$$ 4^{-3} = 4^{6s+3} $$

Solve for s

Since the bases are equal, we equate the exponents:

$$ -3 = 6s + 3 $$

Subtract 3 from both sides:

$$ -6 = 6s $$

Divide by 6:

$$ s = -1 $$

Answer:

• (A) -1 (Correct answer)
• (B) 0
• (C) 1
• (D) no solution