Question

express 64^-3 as an exponent with base 4

Answer 1

Your answer is 4^-9

Hole it helps u :)

Answer 2

64⁻³  =  4⁻⁹ as an exponent with base 4

Given

• 64⁻³

To Find:

• Express 64⁻³  as an exponent with base 4

Laws of exponents

$\begin{align} & {{\text{a}}^{n}}\times {{a}^{-n}}=1\text{ or }{{\text{a}}^{n}}=\frac{1}{{{a}^{n}}} \\ & {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} \\ & \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \\ & {{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}={{a}^{mn}} \\ & {{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}} \\ & \frac{{{a}^{n}}}{{{b}^{n}}}={{\left( \frac{a}{b} \right)}^{n}} \\ & {{a}^{0}}=1 \\ \end{align}$

Step 1 :

Rewrite 64 = 4 × 4 × 4 = 4³

64⁻³  = (4³)⁻³

Step 2 :

Use identity ${{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}={{a}^{mn}}$  and 3 * (-3) = -9

(4³)⁻³ = 4⁻⁹

Hence  64⁻³  =  4⁻⁹ as an exponent with base 4