to view answer.When dealing with problems and expressions involving exponents, it is often useful to simplify them by using the laws of exponents. These “laws” are really just properties of exponents that follow from the definition of an exponent. We will take a look at these properties and see why they make sense:
For the following, let x and y be real numbers or variables, and let m and n be integers.
Property 1:
$$
x^{\large m} x^{\large n} = x^{\large {m + n}}
$$
Why does this make sense?
$$
4^{\large 3} \cdot 4^{\large 5}
$$
Recall that an exponent is simply a repeated multiplication. For instance, $$
4^{\large 3}
$$
is 4 multiplied by itself 3 times: $$
4^{\large 3} = 4 \cdot 4 \cdot 4
$$
Therefore:
$$
4^{\large 3} \cdot 4^{\large 5} = \left( {4 \cdot 4 \cdot 4} \right)\left( {4 \cdot 4 \cdot 4 \cdot 4 \cdot 4} \right) = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 = 4^{\large 8}
$$
So, this is why Property 1 works, and if we understand it, we can just add the exponents:
$$
4^{\large 3} \cdot 4^{\large 5} = 4^{\large {3 + 5}} = 4^{\large 8}
$$
Property 2:
$$
{\Large\frac{{x^{\Large m} }}
{{x^{\Large n} }} = x^{\large {m - n}}}
$$
$$
{\Large\frac{{7^9 }}
{{7^6 }}}
$$
So, again, using the definition of an exponent as a repeated multiplication, we can rewrite this as:
$$
{\Large\frac{{7^9 }}
{{7^6 }} = \frac{{7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7}}
{{7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7}}
}$$
Since these are multiplied together, we can divide out all of the factors that the numerator and the denominator have in common:
$$
{\Large\frac{{7^9 }}
{{7^6 }} = \frac{{\rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot 7 \cdot 7 \cdot 7}}
{{\rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7 \cdot \rlap{--} 7}} = \frac{{7 \cdot 7 \cdot 7}}
{1}} = 7^{\large 3}
$$
So, now we see why subtracting the exponents is the same as dividing out the number of factors that they share, so we can use the property, and get the same result:
$$
{\Large\frac{{7^9 }}
{{7^6 }}} = 7^{\large{9 - 6}} = 7^{\large 3 }
$$
Property 3:
$$
\left( {xy} \right)^{\large m} = x^{\large m} y^{\large m}
$$
$$
\left( {5n} \right)^{\large 3}
$$
We can again use the definition of an exponent to rewrite this as:
$$
\left( {5n} \right)^{\large 3} = 5n \cdot 5n \cdot 5n
$$
By the commutative property, we can multiply these numbers in any order we like, and thus, we can rewrite this as:
$$
\left( {5n} \right)^{\large3} = 5n \cdot 5n \cdot 5n = 5 \cdot 5 \cdot 5 \cdot n \cdot n \cdot n
$$
And, finally, re-writing as exponents, we get:
$$
\left( {5n} \right)^{\large 3} = 5 \cdot 5 \cdot 5 \cdot n \cdot n \cdot n = 5^{\large 3} n^{\large 3}
$$
Thus, Property 3 gives the same result by raising every factor within the parentheses to the same exponent:
$$
\left( {5n} \right)^{\large 3} = 5^{\large 3} n^{\large 3}
$$
Property 4:
$$
{\large \left( {\frac{x}
{y}} \right)}^{\large m} ={\Large \frac{{x^{\Large m} }}
{{y^{\Large m} }}}
$$
$$
{\Large\left( {\frac{2}
{3}} \right)}^{\large 4}
$$
As in the other properties, let’s rewrite this using exponents:
$$
{\Large \left( {\frac{2}
{3}} \right)}^{\large 4} ={\Large \frac{2}
{3} \cdot \frac{2}
{3} \cdot \frac{2}
{3} \cdot \frac{2}
{3}}
$$
When multiplying fractions, we just multiply the numerators together and the denominators together, so we get:
$$
{\Large\left( {\frac{2}
{3}} \right)}^{\large 4} ={\Large \frac{2}
{3} \cdot \frac{2}
{3} \cdot \frac{2}
{3} \cdot \frac{2}
{3} = \frac{{2 \cdot 2 \cdot 2 \cdot 2}}
{{3 \cdot 3 \cdot 3 \cdot 3}}}
$$
Now, re-writing this using exponents, we get:
$$
{\Large\left( {\frac{2}
{3}} \right)}^{\large 4} ={\Large \frac{{2 \cdot 2 \cdot 2 \cdot 2}}
{{3 \cdot 3 \cdot 3 \cdot 3}} = \frac{{2^4 }}
{{3^4 }}}
$$
Thus, we can actually just use Property 4, and “distribute” the exponent to every factor within the fraction:
$$
{\Large\left( {\frac{2}
{3}} \right)}^{\large 4} ={\Large \frac{{2^4 }}
{{3^4 }}}
$$
Property 5:
$$
\left( {x^{\large m} } \right)^{\large n} = x^{\large {m \cdot n} }
$$
$$
\left( {5^{\large 3} } \right)^{\large 2}
$$
First, let’s rewrite the square as multiplication:
$$
\left( {5^{\large 3} } \right)^{\large 2} = \left( {5^{\large 3} } \right)\left( {5^{\large 3 }} \right)
$$
Now, within each parentheses, rewrite the exponents as multiplications:
$$
\left( {5^{\large 3} } \right)^{\large 2} = \left( {5^{\large 3} } \right)\left( {5^{\large 3} } \right) = \left( {5 \cdot 5 \cdot 5} \right)\left( {5 \cdot 5 \cdot 5} \right)
$$
Since these are multiplications, they can be re-written without the parentheses because of the associative property, and we get:
$$
\left( {5^{\large 3} } \right)^{\large 2} = \left( {5 \cdot 5 \cdot 5} \right)\left( {5 \cdot 5 \cdot 5} \right) = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5
$$
Finally, re-writing this as a single exponent, we get:
$$
\left( {5^{\large 3} } \right)^{\large 2} = 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 5^{\large 6}
$$
So, as you can see, since we ended up with 2 factors from the first exponent, and then each of these had 3 factors of 5 within them, we got a total of $$2 \cdot 3 = 6$$ factors of 5. So, we can just use Property 5 and get:
$$
\left( {5^{\large 3} } \right)^{\large 2} = 5^{\large {3 \cdot 2}} = 5^{\large 6 }
$$
Property 6:
$$
x^{\large 0} = 1
$$ when $$
x \ne 0
$$
$$
3^{\large 0} = 1
$$
Why is this true?
Well, what is $${\Large \frac{{3^5 }}{{3^0 }}}$$? According to Property 2, it is $$
3^{\large {5 - 0}} = 3^{\large 5} $$. So, what number can we divide $$
3^{\large 5}$$ by to get $$3^{\large 5} $$? The only possible number is 1. Thus, $$3^{\large 0} $$ must be equal to 1.
Property 7:
$$
{\Large\frac{1}
{{x^n }}} = x^{\large{ - n}}
$$
Why is this true?
From Property 6, we can rewrite 1 as $$x^{\large 0} $$. Now we have: $$
{\Large\frac{{x^0 }}{{x^n }}}$$. Using Property 2, we get:
$$
{\Large \frac{{x^0 }}
{{x^n }}} = x^{\large{0 - n}} = x^{\large{ - n}}
$$
Rewrite with only positive exponents:
$$
{\Large\frac{{3^{ - 4} }}
{{2^{ - 3} }}}
$$
Let’s rewrite the numerator and the denominator using Property 7:
$$
{\huge\frac{{\frac{1}
{{3^{\Large 4} }}}}
{{\frac{1}
{{2^{\Large 3} }}}}} = {\Large\frac{1}
{{3^4 }}} \div {\Large\frac{1}
{{2^3 }}}
$$
Then, dividing fractions, we get:
$$
{\Large\frac{1}{{3^4 }}} \div{\Large \frac{1}{{2^3 }}} = {\Large\frac{1}{{3^4 }}} \cdot{\Large \frac{{2^3 }}{1} }= {\Large\frac{{2^3 }}{{3^4 }}}
$$
So, basically, we see that any negative exponent in the numerator can be moved to the denominator and made positive, and any negative exponent in the denominator can be moved to the numerator and made positive.
Let’s combine some of these rules of exponents in a final example:
Simplify and make all exponents positive:
$$
{\Large\frac{{18x^3 y^7 z^{ - 3} }}{{3y^8 z^{ - 9} x^3 }}}
$$
First, we will simplify the real numbers. We can take a factor of 3 out of 18 and then cancel out the 3’s to get:
$$
{\Large\frac{{18x^3 y^7 z^{ - 3} }}{{3y^8 z^{ - 9} x^3 }}} ={\Large \frac{{\not 3 \cdot 6 \cdot x^3 y^7 z^{ - 3} }}{{\not 3 \cdot y^8 z^{ - 9} x^3 }}} ={\Large \frac{{6 \cdot x^3 y^7 z^{ - 3} }}{{y^8 z^{ - 9} x^3 }}}
$$
Now, let’s move on to the x’s. Using Property 2, we can subtract the exponents of the x’s, to get:
$$
{\Large\frac{{6 \cdot x^3 y^7 z^{ - 3} }}{{y^8 z^{ - 9} x^3 }}} = {\Large\frac{{6y^7 z^{ - 3} }}{{y^8 z^{ - 9} }} \cdot x^{3 - 3}} ={\Large \frac{{6y^7 z^{ - 3} }}{{y^8 z^{ - 9} }}} \cdot x^{\large 0}
$$
And, because of Property 6, we know that $$x^{\large 0} = 1$$. So, now our expression is:
$$
{\Large\frac{{6y^7 z^{ - 3} }}{{y^8 z^{ - 9} }}}
$$
Now, again using Property 2, we can subtract the exponents of the y’s:
$$
{\Large\frac{{6y^7 z^{ - 3} }}{{y^8 z^{ - 9} }}} ={\Large \frac{{6z^{ - 3} }}
{{z^{ - 9} }}} \cdot y^{\large{7 - 8}} ={\Large \frac{{6z^{ - 3} }}{{z^{ - 9} }}} \cdot y^{\large{ - 1}}
$$
By Property 7, we know that a negative exponent means that we can move the y to the bottom and the exponent will be positive 1 instead of -1:
$$
{\Large\frac{{6z^{ - 3} }}{{z^{ - 9} }}} \cdot y^{\large{ - 1}} ={\Large \frac{{6z^{ - 3} }}{{z^{ - 9} y^1 }}} = {\Large\frac{{6z^{ - 3} }}{{z^{ - 9} y^{} }}}
$$
Finally, let’s use Property 2 one more time to subtract the exponents of the z’s:
$$
{\Large\frac{{6z^{ - 3} }}{{z^{ - 9} y^{} }}} ={\Large \frac{6}{y}} \cdot z^{\large{ - 3 - \left( { - 9} \right)}} ={\Large \frac{6}{y}} \cdot z^{\large 6}
$$
So, z has a positive exponent, and therefore should be in the numerator. So, our final, simplified fraction is:
$$
{\Large\frac{{6z^6 }}{y}}
$$
Some practice problems to check your skills:
Simplify the following expressions and make all exponents positive:
1. $$
\left( {3^{\large 2} x} \right)^{\large 6} \left( {3x^{\large 2} } \right)^{\large 2}
$$
Think about it for a moment and then access this link to view answer.
2. $$
\left( {2ab} \right)^{\large { - 2}} \left( {4a^{\large 2} b} \right)^{\large 2}
$$
Think about it for a moment and then access this link to view answer.
3. $$
{\Large\frac{{\left( {3x + 2} \right)^{ - 6} }}
{{\left( {3x + 2} \right)^6 }}}
$$
Think about it for a moment and then access this link to view answer.
4. $${\Large
\frac{{5a^2 b^8 }}
{{24c^5 d^{24} }}}
$$
Think about it for a moment and then access this link to view answer.
5. $$
{\Large\frac{{3\left( {xy^{ - 2} z^4 } \right)^2 }}
{{\left( {3xyz} \right)^{ - 5} }}}
$$
Think about it for a moment and then access this link to view answer.