Solve (2^{2}:2^5+6^5:6^3-3*7+4^0)

$$( 2 ^ { 2 } : 2 ^ { 5 } + 6 ^ { 5 } : 6 ^ { 3 } - 3 \times 7 + 4 ^ { 0 } )$$

Evaluate

$$\frac{129}{8}=16.125$$

Solution Steps

Rewrite 2^{5} as 2^{2}\times 2^{3}. Cancel out 2^{2} in both numerator and denominator.

$$\frac{1}{2^{3}}+\frac{6^{5}}{6^{3}}-3\times 7+4^{0}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 3 from 5 to get 2.

$$\frac{1}{2^{3}}+6^{2}-3\times 7+4^{0}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{1}{8}+6^{2}-3\times 7+4^{0}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{1}{8}+36-3\times 7+4^{0}$$

Add \frac{1}{8} and 36 to get \frac{289}{8}.

$$\frac{289}{8}-3\times 7+4^{0}$$

Multiply 3 and 7 to get 21.

$$\frac{289}{8}-21+4^{0}$$

Subtract 21 from \frac{289}{8} to get \frac{121}{8}.

$$\frac{121}{8}+4^{0}$$

Calculate 4 to the power of 0 and get 1.

$$\frac{121}{8}+1$$

Add \frac{121}{8} and 1 to get \frac{129}{8}.

$$\frac{129}{8}$$

Factor

$$\frac{3 \cdot 43}{2 ^ {3}} = 16\frac{1}{8} = 16.125$$