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Write each of the following products with a single base. Do not simplify further.

  1. ( ( 3 y ) 8 ) 3
  2. ( t 5 ) 7
  3. ( ( g ) 4 ) 4
  1. ( 3 y ) 24
  2. t 35
  3. ( g ) 16
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Using the zero exponent rule of exponents

Return to the quotient rule. We made the condition that m > n so that the difference m n would never be zero or negative. What would happen if m = n ? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.

t 8 t 8 = t 8 t 8 = 1

If we were to simplify the original expression using the quotient rule, we would have

t 8 t 8 = t 8 8 = t 0

If we equate the two answers, the result is t 0 = 1. This is true for any nonzero real number, or any variable representing a real number.

a 0 = 1

The sole exception is the expression 0 0 . This appears later in more advanced courses, but for now, we will consider the value to be undefined.

The zero exponent rule of exponents

For any nonzero real number a , the zero exponent rule of exponents states that

a 0 = 1

Using the zero exponent rule

Simplify each expression using the zero exponent rule of exponents.

  1. c 3 c 3
  2. −3 x 5 x 5
  3. ( j 2 k ) 4 ( j 2 k ) ( j 2 k ) 3
  4. 5 ( r s 2 ) 2 ( r s 2 ) 2

Use the zero exponent and other rules to simplify each expression.


  1. c 3 c 3 = c 3 3 = c 0 = 1

  2. −3 x 5 x 5 = −3 x 5 x 5 = −3 x 5 5 = −3 x 0 = −3 1 = −3

  3. ( j 2 k ) 4 ( j 2 k ) ( j 2 k ) 3 = ( j 2 k ) 4 ( j 2 k ) 1 + 3 Use the product rule in the denominator . = ( j 2 k ) 4 ( j 2 k ) 4 Simplify . = ( j 2 k ) 4 4 Use the quotient rule . = ( j 2 k ) 0 Simplify . = 1

  4. 5 ( r s 2 ) 2 ( r s 2 ) 2 = 5 ( r s 2 ) 2 2 Use the quotient rule . = 5 ( r s 2 ) 0 Simplify . = 5 1 Use the zero exponent rule . = 5 Simplify .
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Simplify each expression using the zero exponent rule of exponents.

  1. t 7 t 7
  2. ( d e 2 ) 11 2 ( d e 2 ) 11
  3. w 4 w 2 w 6
  4. t 3 t 4 t 2 t 5
  1. 1
  2. 1 2
  3. 1
  4. 1
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Using the negative rule of exponents

Another useful result occurs if we relax the condition that m > n in the quotient rule even further. For example, can we simplify h 3 h 5 ? When m < n —that is, where the difference m n is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal.

Divide one exponential expression by another with a larger exponent. Use our example, h 3 h 5 .

h 3 h 5 = h h h h h h h h = h h h h h h h h = 1 h h = 1 h 2

If we were to simplify the original expression using the quotient rule, we would have

h 3 h 5 = h 3 5 =   h −2

Putting the answers together, we have h −2 = 1 h 2 . This is true for any nonzero real number, or any variable representing a nonzero real number.

A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa.

a n = 1 a n and a n = 1 a n

We have shown that the exponential expression a n is defined when n is a natural number, 0, or the negative of a natural number. That means that a n is defined for any integer n . Also, the product and quotient rules and all of the rules we will look at soon hold for any integer n .

The negative rule of exponents

For any nonzero real number a and natural number n , the negative rule of exponents states that

a n = 1 a n

Using the negative exponent rule

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

  1. θ 3 θ 10
  2. z 2 z z 4
  3. ( −5 t 3 ) 4 ( −5 t 3 ) 8
  1. θ 3 θ 10 = θ 3 10 = θ −7 = 1 θ 7
  2. z 2 z z 4 = z 2 + 1 z 4 = z 3 z 4 = z 3 4 = z −1 = 1 z
  3. ( −5 t 3 ) 4 ( −5 t 3 ) 8 = ( −5 t 3 ) 4 8 = ( −5 t 3 ) −4 = 1 ( −5 t 3 ) 4
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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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what is inorganic
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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