What is (-3)^0+(-6)^1+2^4+1^7 ?

The solution to (-3)^0+(-6)^1+2^4+1^7 is 12

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(-3)^0+(-6)^1+2^4+1^7

(- 3)^{0} + (- 6)^{1} + 2^{4} + 1^{7}

12

Solution steps

(- 3)^{0} + (- 6)^{1} + 2^{4} + 1^{7}

Follow the PEMDAS order of operations

Calculate exponents

(- 3)^{0} : 1

(- 3)^{0}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{0} = 3^{0} = 1

= 1

= 1 + (- 6)^{1} + 2^{4} + 1^{7}

Calculate exponents

(- 6)^{1} : - 6

(- 6)^{1}

Apply exponent rule:

(- a)^{n} = - a^{n},

n

is odd

(- 6)^{1} = - 6^{1} = - 6

= - 6

= 1 - 6 + 2^{4} + 1^{7}

Calculate exponents

2^{4} : 16

2^{4}

2^{4} = 16

= 16

= 1 - 6 + 2^{4} + 1^{7}

Calculate exponents

1^{7} : 1

1^{7}

1^{7} = 1

= 1

= 1 - 6 + 2^{4} + 1^{7}

Add and subtract (left to right)

1 - 6 + 2^{4} + 1^{7} : 12

1 - 6 + 2^{4} + 1^{7}

1 - 6 = - 5

= - 5 + 2^{4} + 1^{7}

- 5 + 2^{4} = 11

= 11 + 1^{7}

11 + 1^{7} = 12

= 12

= 12

- 7^{2} - (5^{2} - 4^{2})

- 11 + 1 - 10 - 3

3/2 - 1/3 + 1/4

- 3 (2)^{3}

2 + 1 \times \frac{6}{5}