SIMPLIFYING MONOMIALS WITH NEGATIVE EXPONENTS

Negative Exponent Rule :

To change the negative exponent as positive, we have two ways.

(i) Change the place

(ii) Take the reciprocal.

Base with negative sign :

If base is having negative sign, we have to consider the power. 

• If the power is odd, the result will also have negative sign.
• If power is even, the result will have positive sign.

Write the expression with only positive exponents. Assume all variables represent non zero numbers. Simplify if necessary.

Problem 1 : 

4r³(r⁴)³ / 15(r³)^-2

(a) 4/15r²¹  (b)  4r²¹/15  (c)  4r⁹/15  (d)  4/15r⁹

Solution :

Problem 2 : 

(-4x⁴y^-5)³ (2x^-1y)

(a) -8x⁵/y⁶  (b)  -2x³/y⁴  (c)  -8x^-3/y⁴  (d)  -8x³y⁶

Solution :

Problem 3 : 

[(12x^-5y^-3 z⁴) / (3xy^-3z^-4)]^-1

(a) x⁴/4z⁸  (b)  x⁶/4z⁸  (c)  4x⁶/z⁸  (d)  x⁶y⁶/4z⁸

Solution :

Problem 4 : 

[(xy⁵) / (x³y)]^-2

(a) 1/x⁸ y¹²  (b)  x⁴/y⁸  (c)  1/x⁵y¹¹  (d)  y⁸/x⁴

Solution :

= [(xy⁵) / (x³y)]^-2

= [(x³y) / (xy⁵)]²

By distributing the power, we get

= [ (x³x) / (y⁵ y) ]²

= (x⁴/y⁶)²

= (x⁸/y¹²)

Problem 5 :

Which is the simplified form of (3a²b³c^-2)/(a^-1b²c)³ ?

(a) 3a⁵/b³c⁵  (b)  3ab/c⁵  (c) 3/b²c⁵  (d) 3/ab³c⁵

Solution :

Problem 5 :

A jelly fish emits about 1.25 × 10⁸ particles of light, or photons, in 6.25 × 10⁻⁴ second. How many photons does the jelly fish emit each second? Write your answer in scientific notation and in standard form.

Solution :

Number of particles it emits = 1.25 × 10⁸

Time taken = 6.25 × 10⁻⁴

Number of photons that  jelly fish emit = 6.25 × 10⁻⁴ seconds 

unit rate = (1.25 × 10⁸) / (6.25 × 10⁻⁴)

= (1.25/6.25) x 10⁸ x 10⁴

= 0.2 x 10⁸⁺⁴

= 0.2 x 10¹²

Converting into scientific notation, we get

= 2 x 10^{12 - 1}

= 2 x 10¹¹

Jelly fish emits 2 x 10¹¹ photons per second

Problem 6 :

A microscope magnifies an object 10⁵ times. The length of an object is 10⁻⁷ meter. What is its magnified length?

Solution :

Length of the object = 10⁻⁷ meter

The size magnifies = 10⁵ times

Length of object after it magnifies =  10⁵ (10⁻⁷)

= 10^5-7

= 10^-2

= 1/10²

= 1/100

Problem 7 :

Simplify the expression. Write your answer using only positive exponents.

a)  x^-7

b)  y⁰

c)  9x⁰ y^-3

d)  15c^-8 d⁰

e)  (2^-2 m^-3)/n⁰

f)  (10⁰ r^-11s)/3²

g)  (4^-3 a⁰)/b^-7

h)  p^-8 / 7^-2 q^-9

i)  (2² y^-6)/8^-1 z⁰ x^-7

Solution :

a)  x^-7

= 1/x⁷

b)  y⁰

= 1

c)  9x⁰ y^-3

= 9(1)/y³

= 9/y³

d)  15c^-8 d⁰

= 15c^-8 (1)

To convert the negative exponent to positive,

= 15/c⁸

e)  (2^-2 m^-3)/n⁰

= (2^-2 m^-3)/(1)

= (1/2²) (1/m³)

= (1/4/m³)

f)  (10⁰ r^-11s)/3²

= [(1) r^-11s]/9

= ([1/r¹¹]s) / 9

= s/9r¹¹

g)  (4^-3 a⁰)/b^-7

= (4^-3 (1))/b^-7

= b⁷/4³

= b⁷/64

h)  p^-8 / 7^-2 q^-9

=  7² q⁹ / p⁸ 

=  49 q⁹ / p⁸ 

i)  (2² y^-6)/8^-1 z⁰ x^-7

= (4 y^-6)/8^-1 (1) x^-7

=  (4)(8) /y⁶ x^-7

=  32x⁷/y⁶ 

Problem 8 :

The mass of a grain of sand is about 10^-3  gram. About how many grains of sand are in the bag of sand?

Solution :

= 10 kg x (1000 g/1 kg) x (1 grain of sand/ 10^-3) 

= 10⁴ x 10³

= 10⁴⁺³

= 10⁷  grains of sand.

Problem 9 :

Evaluate and order 5⁰, 5⁴, and 5^-5 from least to greatest

Solution :

5⁰ = 1

5⁴ = 625

5^-5 = 1/5⁵

= 1/3125

Ordering from least to greatest

5^-5, 5⁴, 5⁰

Problem 10 :

A drop of water leaks from a faucet every second. How many liters of water leak from the faucet in 1 hour?

Solution :

1 hour = 3600 seconds

Water leaks from the faucet at a rate of 50^-2 liter per second. Multiply the time by the rate.

3600 ⋅ 50⁻² = 3600 ⋅ 1/50² 

= 3600 ⋅ 1/2500

= 3600/2500 

= 1.44

So, 1.44 liters of water leak from the faucet in 1 hour.