# How to Solve Large Exponents

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As with most problems in basic algebra, solving large exponents requires factoring. If you factor the exponent down until all the factors are prime numbers – a process called prime factorization – you can then apply the power rule of exponents to solve the problem. Additionally, you can break the exponent down by addition rather than multiplication and apply the product rule for exponents to solve the problem. A little practice will help you predict which method will be easiest for the problem you are faced with.

1. ## Find Prime Factors

2. Find the prime factors of the exponent. Example: 624

24 = 2 × 12, 24 = 2 × 2 × 6, 24 = 2 × 2 × 2 × 3

3. ## Apply the Power Rule

4. Use the power rule for exponents to set up the problem. The power rule states: (xa)b = x(a × b)

624 = 6(2 × 2 × 2 × 3) = (((62)2)2)3

5. ## Calculate the Exponents

6. Solve the problem from the inside out.

(((62)2)2)3 = ((362)2)3 = (12962)3 = 16796163 = 4.738 × e18

1. ## Deconstruct the Exponent

2. Break the exponent down into a sum. Make sure the components are small enough to work with as exponents and do not include 1 or 0.

Example: 624

24 = 12 + 12, 24 = 6 + 6 + 6 + 6, 24 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3

3. ## Apply the Product Rule

4. Use the product rule of exponents to set up the problem. The product rule states: xa × xb = x(ab)

624 = 6(3 + 3 + 3 + 3 + 3 + 3 + 3 + 3), 624 = 63 × 63 × 63 × 63 × 63 × 63 × 63 × 63

5. ## Compute the Exponents

6. Solve the problem.

63 × 63 × 63 × 63 × 63 × 63 × 63 × 63 = 216 × 216 × 216 × 216 × 216 × 216 × 216 × 216 = 46656 × 46656 × 46656 × 46656 = 4.738 × e18

#### Things You'll Need

• Pen or pencil
• Paper

#### Tips

• For some problems, a combination of both techniques may make the problem easier. For example: x21 = (x7)3 (power rule), and x7 = x3 × x2 × x2 (product rule). Combining the two, you get: x21 = (x3 × x2 × x2)3

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