Evaluate (4/3)^-8*(4/3)^2 ?

${(\frac{4}{3})}^{- 8} \cdot {(\frac{4}{3})}^{2}$ ?

Step 1

Multiply ${(\frac{4}{3})}^{- 8}$ by ${(\frac{4}{3})}^{2}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(\frac{4}{3})}^{- 8 + 2}$

Add $- 8$ and $2$ .

${(\frac{4}{3})}^{- 6}$

Step 2

Change the sign of the exponent by rewriting the base as its reciprocal.

${(\frac{3}{4})}^{6}$

Step 3

Apply the product rule to $\frac{3}{4}$ .

$\frac{3^{6}}{4^{6}}$

Step 4

Raise $3$ to the power of $6$ .

$\frac{729}{4^{6}}$

Step 5

Raise $4$ to the power of $6$ .

$\frac{729}{4096}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{729}{4096}$

Decimal Form:

$0.17797851 \dots$