(3/8)^-2 = 64/9 show all work

1 answer

To solve \( \left( \frac{3}{8} \right)^{-2} = \frac{64}{9} \), we'll start by applying the negative exponent rule, which states that \( a^{-n} = \frac{1}{a^n} \).

Step 1: Apply the Negative Exponent Rule

\[ \left( \frac{3}{8} \right)^{-2} = \frac{1}{\left( \frac{3}{8} \right)^{2}} \]

Step 2: Compute \( \left( \frac{3}{8} \right)^{2} \)

\[ \left( \frac{3}{8} \right)^{2} = \frac{3^{2}}{8^{2}} = \frac{9}{64} \]

Step 3: Substitute Back into the Original Expression

\[ \left( \frac{3}{8} \right)^{-2} = \frac{1}{\frac{9}{64}} = \frac{64}{9} \]

Conclusion

Thus,

\[ \left( \frac{3}{8} \right)^{-2} = \frac{64}{9} \]

Therefore, the original equation \( \left( \frac{3}{8} \right)^{-2} = \frac{64}{9} \) is verified to be true.