Question

solve the equation.
16^{3x - 3} = 64^{x + 2}

Explanation:

Step1: Express bases as powers of 2

$16=2^4$, $64=2^6$, so substitute:
$$(2^4)^{3x-3}=(2^6)^{x+2}$$

Step2: Simplify exponents via power rule

Use $(a^m)^n=a^{m \cdot n}$:
$$2^{4(3x-3)}=2^{6(x+2)}$$

Step3: Set exponents equal (same base)

Since bases are equal, exponents match:
$$4(3x-3)=6(x+2)$$

Step4: Expand both sides

Distribute coefficients:
$$12x-12=6x+12$$

Step5: Isolate x terms

Subtract $6x$ from both sides:
$$6x-12=12$$

Step6: Solve for x

Add 12, then divide by 6:
$$6x=24 \implies x=\frac{24}{6}=4$$

Answer:

$x=4$