Question

question 7 (multiple choice worth 1 points)
(05.01r mc)
solve $64^x = 16^{x-1}$.
$x = -2$
$x = -1$
$x = \frac{-1}{4}$
$x = \frac{-1}{3}$

Explanation:

Step1: Rewrite bases as powers of 4

$64 = 4^3$, $16 = 4^2$, so:
$$(4^3)^x = (4^2)^{x-1}$$

Step2: Simplify exponents

Use exponent rule $(a^m)^n=a^{mn}$:
$$4^{3x} = 4^{2(x-1)}$$

Step3: Set exponents equal

Since bases are equal, exponents are equal:
$$3x = 2(x-1)$$

Step4: Expand right-hand side

$$3x = 2x - 2$$

Step5: Solve for x

Subtract $2x$ from both sides:
$$3x - 2x = -2 \implies x = -2$$

Answer:

A. $x=-2$