Examples with solutions for Powers and Roots: Solving the equation

Exercise #1

Solve:

524+33 5^2\cdot4+3^3

Video Solution

Step-by-Step Solution

Remember that according to the order of arithmetic operations, exponents precede multiplication and division, which precede addition and subtraction (and parentheses always precede everything).

So first calculate the values of the terms with exponents and then subtract the results:

524+33=254+27=100+27=127 5^2\cdot4+3^3 =25\cdot4+27=100+27=127 Therefore, the correct answer is option B.

Answer

127

Exercise #2

Sovle:

32+33 3^2+3^3

Video Solution

Step-by-Step Solution

Remember that according to the order of operations, exponents precede multiplication and division, which precede addition and subtraction (and parentheses always precede everything).

So first calculate the values of the terms in the power and then subtract between the results:

32+33=9+27=36 3^2+3^3 =9+27=36 Therefore, the correct answer is option B.

Answer

36

Exercise #3

What is the answer to the following?

3233 3^2-3^3

Video Solution

Step-by-Step Solution

Remember that according to the order of operations, exponents come before multiplication and division, which come before addition and subtraction (and parentheses always before everything),

So first calculate the values of the terms in the power and then subtract between the results:

3233=927=18 3^2-3^3 =9-27=-18 Therefore, the correct answer is option A.

Answer

18 -18

Exercise #4

Solve:

1642331 \sqrt{16}\cdot4^2-3^3\cdot\sqrt{1}

Step-by-Step Solution

Begin by evaluating the square roots: 16=4 \sqrt{16} = 4 and 1=1 \sqrt{1} = 1 .

Substitute these back into the expression:

442331 4\cdot4^2-3^3\cdot1

Calculate each term:

  • 42=16 4^2 = 16 , so 416=64 4\cdot16 = 64

  • 33=27 3^3 = 27 , so 271=27 27\cdot1 = 27

Subtract the second result from the first:

6427=37 64 - 27 = 37

Answer

60

Exercise #5

Solve:

442521 \sqrt{4}\cdot4^2-5^2\cdot\sqrt{1}

Video Solution

Step-by-Step Solution

We simplify each term according to the order from left to right:

4=2 \sqrt{4}=2

42=4×4=16 4^2=4\times4=16

52=5×5=25 5^2=5\times5=25

1=1 \sqrt{1}=1

Now we rearrange the exercise accordingly:

2×1625×1 2\times16-25\times1

Since there are two multiplication operations in the exercise, according to the order of operations we start with them and then subtract.

We put the two multiplication exercises in parentheses to avoid confusion during the solution, and solve from left to right:

(2×16)(25×1)=3225=7 (2\times16)-(25\times1)=32-25=7

Answer

7

Exercise #6

Solve:

932+234 \sqrt{9}\cdot3^2+2^3\cdot\sqrt{4}

Step-by-Step Solution

First, we need to evaluate the square roots: 9=3 \sqrt{9} = 3 and 4=2 \sqrt{4} = 2 .

Substitute these values back into the expression:

332+232 3\cdot3^2+2^3\cdot2

Calculate each term separately:

  • 32=9 3^2 = 9 , so 39=27 3\cdot9 = 27

  • 23=8 2^3 = 8 , so 82=16 8\cdot2 = 16

Add these values together:

27+16=33 27 + 16 = 33

Answer

33

Exercise #7

Which of the following is equivalent to 1000 100^0 ?

Video Solution

Step-by-Step Solution

Let's solve the problem step by step using the Zero Exponent Rule, which states that any non-zero number raised to the power of 0 is equal to 1.


  • Consider the expression: 1000 100^0 .
  • According to the Zero Exponent Rule, if we have any non-zero number, say a a , then a0=1 a^0 = 1 .
  • Here, a=100 a = 100 which is clearly a non-zero number, so following the rule, we find that:
  • 1000=1 100^0 = 1 .

Therefore, the expression 1000 100^0 is equivalent to 1.

Answer

1