SAT Practice Question
If \(b^3{\cdot}(b^4)\strut^2=b^x\), what is the value of \(x\)? a) \(9\) b) \(11\) c) \(18\) d) \(19\) Source: Khan Academy |
In the Polynomials Unit, we talked about writing equivalent expressions using properties of exponents. We are now going to extend that work to rational exponents. Let’s first do a quick summary of the properties of exponents:
Product of Powers: \(\large{x^a\cdot x^b=x^{\left(a+b\right)}}\)
Power of a Power: \(\large\left(x^a\right)^b=x^{ab}\)
Quotient of Powers: \(\Large{\frac{x^a}{x^b}}=\large{x^{\left(a-b\right)}}\)
Power of a Product: \(\large{\left(x^a\cdot y^b\right)^c=x^{ac}\cdot y^{bc}}\)
Power of a Quotient: \(\left(\Large{\frac{x^a}{y^b}}\right)^c=\Large{\frac{x^{ac}}{y^{bc}}}\)
Zero Exponents: \(x^0=1\)
Negative Exponents: \(\large{x^{-a}}=\Large{\frac{1}{x^a}}\)
Power of a Power: \(\large\left(x^a\right)^b=x^{ab}\)
Quotient of Powers: \(\Large{\frac{x^a}{x^b}}=\large{x^{\left(a-b\right)}}\)
Power of a Product: \(\large{\left(x^a\cdot y^b\right)^c=x^{ac}\cdot y^{bc}}\)
Power of a Quotient: \(\left(\Large{\frac{x^a}{y^b}}\right)^c=\Large{\frac{x^{ac}}{y^{bc}}}\)
Zero Exponents: \(x^0=1\)
Negative Exponents: \(\large{x^{-a}}=\Large{\frac{1}{x^a}}\)
A power that involves a rational exponents is in the form \(x\strut^{\frac{a}{b}}\) where \(a\) and \(b\) are integers. Let’s look at a few examples with actual numbers so you can see what happens with these types of exponents.
Let’s start with the equation \(x^3=8\). You know that the value of \(x\) that makes that true is \(x=2\). Suppose we raise each side of the equation to the \(\frac{1}{3}\) power and let’s see what happens.
Let’s start with the equation \(x^3=8\). You know that the value of \(x\) that makes that true is \(x=2\). Suppose we raise each side of the equation to the \(\frac{1}{3}\) power and let’s see what happens.
\(\begin{align}x^3&=8\\
\left(x^3\right)^{\frac{1}{3}}&=8\strut^{\frac{1}{3}}\ & \ &\text{Raise both sides to the}\ \frac{1}{3}\ \text{power}\\
x&=8\strut^{\frac{1}{3}}\ & \ &\text{Use the power of a power property to multiply the exponents}\end{align}\)
\left(x^3\right)^{\frac{1}{3}}&=8\strut^{\frac{1}{3}}\ & \ &\text{Raise both sides to the}\ \frac{1}{3}\ \text{power}\\
x&=8\strut^{\frac{1}{3}}\ & \ &\text{Use the power of a power property to multiply the exponents}\end{align}\)
Since we also said at the beginning that \(x=2\), then \(8^{\frac{1}{3}}\) must also be equal to \(2\). This same pattern would produce the following statements as well.
\(25\strut^{\frac{1}{2}}=5\)
\(81\strut^{\frac{1}{4}}=3\)
\(\left(-27\right)^{\frac{1}{3}}=-3\)
To generalize these results, raising a number to the \(\frac{1}{n}\) power is the same as taking the nth root of that number. So \(\large{x\strut^{\frac{1}{n}}=\sqrt[n]{x}}\).
What happens if the numerator is not \(1\)? Let’s look at an expression like \(25\strut^{\frac{3}{2}}\). We can use the power of a power property in reverse to rewrite that expression as \(\left(25\strut^{\frac{1}{2}}\right)^3\) since \(\frac{1}{2}\cdot3=\frac{3}{2}\). Based on what we just showed above \(25\strut^{\frac{1}{2}}=5\), so the expression would then be \(5^3\) which is \(125\).
The denominator of the rational exponent was the root we took and the numerator became the power we raised it to. To generalize these results, we would have this statement: \(\large{x\strut^{\frac{a}{b}}=\left(\sqrt[b]{x}\right)^a}\).
A quick note about dealing with negative numbers. In the Quadratics and Polynomials units, we worked often with imaginary numbers when we took the square root of a negative number. As those even powers become larger than \(2\), it becomes much more difficult to work with imaginary numbers and you will do that in PreCalculus. For this unit, we are only going to work with real numbers. If, when solving an equation, we get to a situation where we are trying to take an even root of a negative number, we will simply say that it is not a real number or that there are no real solutions.
Use the table below to help with powers of numbers. You will become very comfortable with these after working through examples and hopefully won’t need to reference the chart much. There are also flashcards in the Target A Quick Quiz for you to quiz yourself to make sure you know these powers.
\(25\strut^{\frac{1}{2}}=5\)
\(81\strut^{\frac{1}{4}}=3\)
\(\left(-27\right)^{\frac{1}{3}}=-3\)
To generalize these results, raising a number to the \(\frac{1}{n}\) power is the same as taking the nth root of that number. So \(\large{x\strut^{\frac{1}{n}}=\sqrt[n]{x}}\).
What happens if the numerator is not \(1\)? Let’s look at an expression like \(25\strut^{\frac{3}{2}}\). We can use the power of a power property in reverse to rewrite that expression as \(\left(25\strut^{\frac{1}{2}}\right)^3\) since \(\frac{1}{2}\cdot3=\frac{3}{2}\). Based on what we just showed above \(25\strut^{\frac{1}{2}}=5\), so the expression would then be \(5^3\) which is \(125\).
The denominator of the rational exponent was the root we took and the numerator became the power we raised it to. To generalize these results, we would have this statement: \(\large{x\strut^{\frac{a}{b}}=\left(\sqrt[b]{x}\right)^a}\).
A quick note about dealing with negative numbers. In the Quadratics and Polynomials units, we worked often with imaginary numbers when we took the square root of a negative number. As those even powers become larger than \(2\), it becomes much more difficult to work with imaginary numbers and you will do that in PreCalculus. For this unit, we are only going to work with real numbers. If, when solving an equation, we get to a situation where we are trying to take an even root of a negative number, we will simply say that it is not a real number or that there are no real solutions.
Use the table below to help with powers of numbers. You will become very comfortable with these after working through examples and hopefully won’t need to reference the chart much. There are also flashcards in the Target A Quick Quiz for you to quiz yourself to make sure you know these powers.
\(x\) |
\(x^2\) |
\(x^3\) |
\(x^4\) |
\(x^5\) |
\(x^6\) |
\(1\) |
\(1\) |
\(1\) |
\(1\) |
\(1\) |
\(1\) |
\(2\) |
\(4\) |
\(8\) |
\(16\) |
\(32\) |
\(64\) |
\(3\) |
\(9\) |
\(27\) |
\(81\) |
\(243\) |
\(729\) |
\(4\) |
\(16\) |
\(64\) |
\(256\) |
\(1024\) |
\(4096\) |
\(5\) |
\(25\) |
\(125\) |
\(625\) |
\(3125\) |
\(15625\) |
\(6\) |
\(36\) |
\(216\) |
\(1296\) |
\(7776\) |
\(46656\) |
\(7\) |
\(49\) |
\(343\) |
\(2401\) |
\(16807\) |
\(117649\) |
\(8\) |
\(64\) |
\(512\) |
\(4096\) |
\(32768\) |
\(262144\) |
\(9\) |
\(81\) |
\(729\) |
\(6561\) |
\(59049\) |
\(531441\) |
Example 1: Simplify each expression in the real numbers.
a) \(81\strut^{\frac{3}{4}}\)
\(\begin{align}&{\left(\sqrt[4]{81}\right)^{\small{3}}}\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\text{1) The denominator of the exponent is the root and the numerator is the power that the}\\
&\ &\ \ \ \ \ \ \ \ \ \ \ \ &\text{ root is being raised to.}\\
&3^3\ &\ \ \ \ \ \ \ \ \ \ \ \ &\text{2) The fourth root of 81 is 3.}\\
&27\ &\ \ \ \ \ \ \ \ \ \ \ \ &\text{3) Simplify.}\end{align}\)
&\ &\ \ \ \ \ \ \ \ \ \ \ \ &\text{ root is being raised to.}\\
&3^3\ &\ \ \ \ \ \ \ \ \ \ \ \ &\text{2) The fourth root of 81 is 3.}\\
&27\ &\ \ \ \ \ \ \ \ \ \ \ \ &\text{3) Simplify.}\end{align}\)
b) \(\left(-36\right)^{\frac{3}{2}}\)
\(\begin{align}&\left(\sqrt{-36}\right)^3\ & \ &\text{1) The denominator of the exponent is the root and the numerator is the power that the root}\\
&\ & \ &\text{is being raised to.}\\
&\text{Not a Real Number}\ & \ &\text{2) Since taking the square root of -36 would produce an imaginary number,}\\
&\ & \ &\text{we will stop and just say it is not a real number.}\end{align}\)
&\ & \ &\text{is being raised to.}\\
&\text{Not a Real Number}\ & \ &\text{2) Since taking the square root of -36 would produce an imaginary number,}\\
&\ & \ &\text{we will stop and just say it is not a real number.}\end{align}\)
c) \(\left(-8\right)\strut^{-\frac{5}{3}}\)
\(\begin{align}&{\frac{1}{{\left(-8\right)}^{\frac{5}{3}}}}\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\text{1) Rewrite using the Negative Exponent property.}\\
&{\frac{1}{\left(\sqrt[3]{-8}\right)^5}}\ & \ &\text{2) The denominator of the exponent is the root and the numerator is the power that the}\\
&\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\text{root is being raised to.}\\
&{\frac{1}{\left(-2\right)^5}}\ & \ &\text{3) The cube root of -8 is -2.}\\
&{-\frac{1}{32}}\ & \ &\text{4) A negative number raised to an odd power is still negative.}\end{align}\)
&{\frac{1}{\left(\sqrt[3]{-8}\right)^5}}\ & \ &\text{2) The denominator of the exponent is the root and the numerator is the power that the}\\
&\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\text{root is being raised to.}\\
&{\frac{1}{\left(-2\right)^5}}\ & \ &\text{3) The cube root of -8 is -2.}\\
&{-\frac{1}{32}}\ & \ &\text{4) A negative number raised to an odd power is still negative.}\end{align}\)
d) \(\large{\left(\frac{27}{125}\right)^{-\frac{2}{3}}}\)
Quick Check
Simplify each expression in the real numbers.
1) \(81\strut^{-\frac{1}{2}}\)
2) \(\left(-\frac{32}{243}\right)^{-\frac{2}{5}}\)
3) \(\left(4\right)^{\frac{3}{2}}\)
Quick Check Solutions
Simplify each expression in the real numbers.
1) \(81\strut^{-\frac{1}{2}}\)
2) \(\left(-\frac{32}{243}\right)^{-\frac{2}{5}}\)
3) \(\left(4\right)^{\frac{3}{2}}\)
Quick Check Solutions
Now that we know how to evaluate using rational exponents, we can now work with these when the base is both a number and a variable. The properties used here will be the same as Target A in the Polynomials Unit and referenced at the beginning of this target.
Example 2: \(5x\strut^{\frac{2}{3}}\cdot7x\strut^{-\frac{1}{4}}\).
\(\begin{align}&5x\strut^{\frac{2}{3}}\cdot{\dfrac{7}{x^{\frac{1}{4}}}}\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\text{1) Use the Negative Exponent Property to rewrite with positive exponents.}\\
&{\dfrac{5x\strut^{\frac{2}{3}}\cdot7}{x\strut^{\frac{1}{4}}}}\ & \ &\text{2) Multiply.}\\
&35x\strut^{{\frac{2}{3}-\frac{1}{4}}}\ & \ &\text{3) When we divide like bases, exponents are subtracted.}\\
&35x\strut^{{\frac{8}{12}-\frac{3}{12}}}\ & \ &\text{4) Find common denominators.}\\
&35x\strut^{\frac{5}{12}}\ & \ &\text{5) Subtracting fractions means subtract numerators and rewrite denominator}\\
& & \ &\text{(simplify fractions to lowest terms if necessary).}\end{align}\)
&{\dfrac{5x\strut^{\frac{2}{3}}\cdot7}{x\strut^{\frac{1}{4}}}}\ & \ &\text{2) Multiply.}\\
&35x\strut^{{\frac{2}{3}-\frac{1}{4}}}\ & \ &\text{3) When we divide like bases, exponents are subtracted.}\\
&35x\strut^{{\frac{8}{12}-\frac{3}{12}}}\ & \ &\text{4) Find common denominators.}\\
&35x\strut^{\frac{5}{12}}\ & \ &\text{5) Subtracting fractions means subtract numerators and rewrite denominator}\\
& & \ &\text{(simplify fractions to lowest terms if necessary).}\end{align}\)
Example 3: \(\left(11^4\cdot16^3\right)^{\frac{1}{4}}\).
Example 4: \(\Large{\frac{20x\strut^{\frac{7}{5}}y\strut^{\frac{3}{4}}}{25xy\strut^{\frac{1}{9}}}}\)