Last modified on October 26th, 2024

chapter outline

 

Simplifying Exponents

Simplifying exponents means reducing the algebraic expressions with exponents to their simplest form. We use certain rules to simplify those expressions.

With the Same Base

When simplifying exponents with the same base, we follow the rules:

  • xm × xn = xm + n 
  • xm ÷ xn = xm – n 

Let us simplify y7 × y3 

= (y × y × y × y × y × y × y) × (y × y × y), which leaves a total of 10 y’s.

Thus, the simplified expression becomes y10 

However, using the product law, it can be easily simplified 

y7 × y3 = y7 + 3 = y10 

With Different Bases

There can be two possible cases when multiplying or dividing exponents with different bases. 

Different Bases and the Same Power

When simplifying exponents with different bases and the same power, we follow the given rules:

Here, we follow the rules:

  • xm × ym = (xy)m 
  • xm ÷ ym = (x ÷ y)m 

Let us simplify p4 × q4 

= (p × p × p × p) × (q × q × q × q)

= (p × q) × (p × q) × (p × q) × (p × q), which leaves a total of 4 (p × q)’s.

Thus, the simplified expression becomes (p × q)4 

However, using the power of product law, it is easily simplified as: 

p4 × q4 = (p × q)4 

Different Bases and Different Powers

When simplifying exponents with different bases and different powers, we follow the given rules:

  • xm × yn = xmyn 
  • xm ÷ yn = ${\dfrac{x^{m}}{y^{n}}}$

Let us simplify 34 ÷ 72 

= (3 × 3 × 3 × 3) ÷ (7 × 7)

= ${\dfrac{81}{49}}$

Thus, the simplified expression becomes ${\dfrac{81}{49}}$ 

Simplifying Fractional Exponents

When simplifying fractional exponents, we follow the following rules:

  • ${x^{\dfrac{m}{n}}\times x^{\dfrac{p}{q}}=x^{\left( \dfrac{m}{n}+\dfrac{p}{q}\right) }}$
  • ${x^{\dfrac{m}{n}}\times y^{\dfrac{m}{n}}=\left( x\times y\right) ^{\dfrac{m}{n}}}$
  • ${x^{\dfrac{m}{n}}\times y^{\dfrac{m}{n}}=x^{\dfrac{m}{n}}y^{\dfrac{m}{n}}}$
  • ${x^{\dfrac{m}{n}}\div x^{\dfrac{p}{q}}=x^{\left( \dfrac{m}{n}-\dfrac{p}{q}\right) }}$
  • ${x^{\dfrac{m}{n}}\div y^{\dfrac{m}{n}}=\left( \dfrac{x}{y}\right) ^{\dfrac{m}{n}}}$
  • ${x^{\dfrac{m}{n}}\div y^{\dfrac{p}{q}}=\dfrac{x^{\dfrac{m}{n}}}{y^{\dfrac{p}{q}}}}$

Let us simplify ${8^{\dfrac{2}{3}}\div 4^{\dfrac{3}{2}}}$

= ${\dfrac{8^{\dfrac{2}{3}}}{4^{\dfrac{3}{2}}}}$

= ${\dfrac{\left( \sqrt[3] {8}\right) ^{2}}{\left( \sqrt{4}\right) ^{3}}}$

= ${\dfrac{\left( 2\right) ^{2}}{\left( 2\right) ^{3}}}$

= ${2^{2-3}}$

= ${2^{-1}}$

= ${\dfrac{1}{2}}$

Thus, the simplified expression is ${\dfrac{1}{2}}$

Simplifying Negative Exponents

When simplifying fractional exponents, we follow the following rules:

  • ${x^{-m}\times x^{-n}=x^{-\left( m+n\right) }=\dfrac{1}{x^{m+n}}}$
  • ${x^{-m}\times y^{-m}=\left( x\times y\right) ^{-m}=\dfrac{1}{\left( x\times y\right) ^{m}}}$
  • ${x^{-m}\times y^{-n}=x^{-m}y^{-n}=\dfrac{1}{x^{-m}y^{-n}}}$
  • ${x^{-m}\div x^{-n}=x^{-\left( m-n\right) }=\dfrac{1}{x^{m-n}}}$
  • ${x^{-m}\div y^{-m}=\left( \dfrac{x}{y}\right) ^{-m}=\left( \dfrac{y}{x}\right) ^{m}}$
  • ${x^{-m}\div y^{-n}=\dfrac{x^{-m}}{y^{-n}}=\dfrac{y^{n}}{x^{m}}}$

Let us simplify 43 × 4-1 

= 43 + (-1) 

= 42 

Thus, the simplified expression is 42 

Solved Examples

Evaluate: x4y0 × (2x-2y5)

Solution:

Here, x4y0 × (2x-2y5)
= 2x4 + (-2)y0 + 5 (using product law)
= 2x2y5

Simplify the rational expression: ${\dfrac{a^{2}b^{-3}}{b^{-4}c^{2}d}}$

Solution:

Here, ${\dfrac{a^{2}b^{-3}}{b^{-4}c^{2}d}}$
= ${\dfrac{a^{2}b^{-3-\left( -4\right) }}{c^{2}d}}$ (using quotient law)
= ${\dfrac{a^{2}b}{c^{2}d}}$

Simplify: ${\dfrac{\left( 3x^{2}y^{3}\right) ^{2}}{x^{2}y^{-1}}}$

Solution:

Here, ${\dfrac{\left( 3x^{2}y^{3}\right) ^{2}}{x^{2}y^{-1}}}$
= ${\dfrac{3^{2}\left( x^{2}\right) ^{2}\left( y^{3}\right) ^{2}}{x^{2}y^{-1}}}$ (using power of product law)
= ${\dfrac{9x^{4}y^{6}}{x^{2}y^{-1}}}$ (using power of power law)
= ${9x^{4-2}y^{6-\left( -1\right) }}$ (using quotient law)
= ${9x^{2}y^{7}}$

Simplify the expression with negative exponents: ${\dfrac{m^{-3}n^{5}}{m^{2}n^{-4}}}$

Solution:

Here, ${\dfrac{m^{-3}n^{5}}{m^{2}n^{-4}}}$
= ${m^{-3-2}n^{5-\left( -4\right) }}$ (using quotient law)
= ${m^{-5}n^{9}}$
= ${\dfrac{n^{9}}{m^{5}}}$ (using law of negative exponents)

Solve: ${\left( \dfrac{3p^{2}q}{5p^{3}q^{2}}\right) ^{2}}$

Solution:

Here, ${\left( \dfrac{3p^{2}q}{5p^{3}q^{2}}\right) ^{2}}$
= ${\left( \dfrac{3}{5}p^{2-3}q^{1-2}\right) ^{2}}$ (using quotient law)
= ${\left( \dfrac{3}{5}p^{-1}q^{-1}\right) ^{2}}$
= ${\left( \dfrac{3}{5pq}\right) ^{2}}$
= ${\dfrac{3^{2}}{5^{2}p^{2}q^{2}}}$ (power of quotient law)
= ${\dfrac{9}{25p^{2}q^{2}}}$

Last modified on October 26th, 2024