Solve 3^{2}*2^2+(3^6/3^4)^0=37

$$3 ^ { 2 } \cdot 2 ^ { 2 } + ( \frac { 3 ^ { 6 } } { 3 ^ { 4 } } ) ^ { 0 } = 37$$

Verify

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 4 from 6 to get 2.

$$3^{2}\times 2^{2}+\left(3^{2}\right)^{0}=37$$

To raise a power to another power, multiply the exponents. Multiply 2 and 0 to get 0.

$$3^{2}\times 2^{2}+3^{0}=37$$

Calculate 3 to the power of 2 and get 9.

$$9\times 2^{2}+3^{0}=37$$

Calculate 2 to the power of 2 and get 4.

$$9\times 4+3^{0}=37$$

Multiply 9 and 4 to get 36.

$$36+3^{0}=37$$

Calculate 3 to the power of 0 and get 1.

$$36+1=37$$

Add 36 and 1 to get 37.

$$37=37$$

Compare 37 and 37.

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