The problem asks us to evaluate the given exponential expressions: $3^3$, $2^5$, $7^4$, $(\frac{2}{7})^3$, and $(\frac{1}{6})^4$.

ArithmeticExponentsOrder of OperationsFractionsPowers
2025/3/31

1. Problem Description

The problem asks us to evaluate the given exponential expressions: 333^3, 252^5, 747^4, (27)3(\frac{2}{7})^3, and (16)4(\frac{1}{6})^4.

2. Solution Steps

* Evaluate 333^3:
33=3×3×3=9×3=273^3 = 3 \times 3 \times 3 = 9 \times 3 = 27.
* Evaluate 252^5:
25=2×2×2×2×2=4×4×2=16×2=322^5 = 2 \times 2 \times 2 \times 2 \times 2 = 4 \times 4 \times 2 = 16 \times 2 = 32.
* Evaluate 747^4:
74=7×7×7×7=49×49=24017^4 = 7 \times 7 \times 7 \times 7 = 49 \times 49 = 2401.
* Evaluate (27)3(\frac{2}{7})^3:
(27)3=2373=2×2×27×7×7=8343(\frac{2}{7})^3 = \frac{2^3}{7^3} = \frac{2 \times 2 \times 2}{7 \times 7 \times 7} = \frac{8}{343}.
* Evaluate (16)4(\frac{1}{6})^4:
(16)4=1464=16×6×6×6=136×36=11296(\frac{1}{6})^4 = \frac{1^4}{6^4} = \frac{1}{6 \times 6 \times 6 \times 6} = \frac{1}{36 \times 36} = \frac{1}{1296}.

3. Final Answer

33=273^3 = 27
25=322^5 = 32
74=24017^4 = 2401
(27)3=8343(\frac{2}{7})^3 = \frac{8}{343}
(16)4=11296(\frac{1}{6})^4 = \frac{1}{1296}