$$
\begin{array}{|c|c|c|}
\hline
\textbf{Rule} & \textbf{Arithmetic example} & \textbf{Algebraic example} \\ \hline
(a^m) (a^n) = a^{m+n} & (5^{3})(5^5) = 5^{3+5}=5^8 & (x^6)(x^{-4})=x^{6+(-4)} = x^2 \\ \hline
(a^m)^n = a^{mn} = (a^n)^m &(2^2)^3=2^{6}=64 &(3z^2)^3=(3^3)(z^2)^3=27z^{6} \\ \hline
a^{-n} = \dfrac{1}{a^n} \quad (a \neq 0)& 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8} & x^{-3} = \dfrac{1}{x^3} \\ \hline
\dfrac{a^m}{a^n} = a^{m-n} & \dfrac{7^{8}}{7^5} = 7^{8-5}=7^3 & \dfrac{x^5}{x^{-4}}=x^{5-(-4)} = x^9\\ \hline
a^0=1 & (-5)^0=1 & x^0=1 \quad (x \neq 0) \\ \hline
(a \times b)^n = (a^n)(b^n) & (2\times 5)^6=(2^6)(5^6) =10^6 & (2x)^3 = (2)^3(x)^3= 8x^3\\ \hline
\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n} & \left(\dfrac{3}{2}\right)^3 = \dfrac{3^3}{2^3}=\dfrac{27}{8} & \left(\dfrac{2x}{3y^{2}}\right)^3 = \dfrac{2^3 x^3}{3^{3} (y^{2})^3}=
\dfrac{8x^3}{27 y^{6}}\\ \hline
\end{array}
$$
Exponents: Algebraic Manipulation
You will be expected to rewrite and manipulate algebraic expressions containing exponent terms. Here I list some examples of common manipulations:
$$8^x = (2^3)^x = 2^{3x}$$ $$27^x = (3^3)^x = 3^{3x}$$ $$3^{6x} = (3^2)^{3x} = 9^{3x}$$ $$3^{6x} = (3^3)^{2x} = 27^{2x}$$ $$9(3^x) = (3^2)(3^x) = 3^{x+2} \quad \text{Note:} \quad 9(3^x) \neq 27^{x}$$ $$8(2^x) = (2^3)(2^x) = 2^{x+3} \quad \text{Note:} \quad 8(2^x) \neq 16^{x}$$ $$\dfrac{3^{x+1}}{3} =\dfrac{3^{x+1}}{3^1}= 3^{(x+1)-1} = 3^x \quad \text{Note:} \quad \dfrac{3^{x+1}}{3} \neq x+1$$ $$ \dfrac{9^x}{27^y} = \dfrac{(3^2)^x}{(3^3)^y} = \dfrac{3^{2x}}{3^{3y}} = 3^{2x-3y} $$
Exponents: Adding and Subtracting Terms
You will be asked to simplify exponent expressions such as $2^{8}+2^{8}$. The most common mistake students make is to add the exponents, $2^8 + 2^8 \neq 2^{16}$, instead $2^8 + 2^8 = 2^{9}$. There is no general exponent rule when adding powers of numbers that have the same base, however, there are cases where simplification is possible using other rules of arithmetic. If you see a question on the SAT that asks you to add exponent terms with the same bases, the best approach is to factor the largest common term which will almost always lead to simplification. For example:
$$2^8 + 2^8 = 2^8(1 + 1) = 2^8(2) = 2^8(2^1) = 2^{8+1} = 2^9$$
Additional Examples
$$ 2^4 + 2^4 + 2^4 + 2^4 = 2^4(1 + 1 + 1 + 1) = 2^4(4) = 2^4(2^2) = 2^{4+2} = 2^6 $$ $$ 2^{20} – 3(2^{18}) = 2^{18}(2^2 – 3) = 2^{18}(4-3) =2^{18} $$ $$ 27^{\frac{5}{3}} – 8(27) = 27(27^{\frac{5}{3} – 1} – 8) = 27(27^{\frac{2}{3}} – 8)
= 27[ (3^3)^{\frac{2}{3}} – 8] = 27( 3^2 – 8) = 27 $$ $$ 2^{x} + 2^{x+1} = 2^x + (2^{x})(2^1) = 2^{x}(1+2) =3(2^x)$$
Radicals and Fractional Exponents
- A square root of a number $n$ is a number that, when squared, is equal to $n$ or $(\sqrt{n})^2 = n$.
- An exponent of $\dfrac{1}{2}$ is the same as taking the square root of a number, $\sqrt{n} = n^{\frac{1}{2}}.$
- A cube root of a number $n$ is a number that, when cubed, is equal to $n$ or $(\sqrt[3]{n})^3 = n$.
- An exponent of $\dfrac{1}{3}$ is the same as taking the cube root of a number, $\sqrt[3]{n} = n^{\frac{1}{3}}$.
$$
\begin{array}{|c|c|}
\hline
\textbf{Rule} & \textbf{Example} \\ \hline
(\sqrt{a})^2 = a \quad (a>0) & (\sqrt{7})^2 = 7 \\ \hline
\sqrt{a^2} = a \quad (a>0) & \sqrt{7^2} = 7 \\ \hline
\sqrt{a \times b} = \sqrt{a} \times \sqrt{b} & \sqrt{6}\sqrt{8}=\sqrt{48} = \sqrt{(16)(3)} = \sqrt{16}\sqrt{3} = 4\sqrt{3} \\ \hline
\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} & \dfrac{\sqrt{6}}{\sqrt{24}} = \sqrt{\dfrac{6}{24}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2} \\ \hline
a^{\frac{m}{n}}= (a^{\frac{1}{n}})^m = (a^m)^{\frac{1}{n}}= (\sqrt[n]{a})^m=\sqrt[n]{a^m} & 27^{\frac{2}{3}} =
(27^2)^{\frac{1}{3}}=(27^{\frac{1}{3}})^2 = (\sqrt[3]{27})^2= (3)^2=9\\ \hline
\end{array}
$$
Here is a list of common exponent manipulations that you can expect on the SAT math section:
$$ \sqrt[6]{27} = 27^{\frac{1}{6}} = (3^3)^{\frac{1}{6}} = 3^{\frac{3}{6}} = 3^{\frac{1}{2}} = \sqrt{3}$$ $$ 64^{-\frac{2}{3}} = (2^6)^{-\frac{2}{3}} = 2^{6 \times -\frac{2}{3}} = 2^{-4} = \frac{1}{2^4} = \frac{1}{16} $$ $$a^{\frac{3}{4}} = (a^{\frac{1}{4}})^3 = (\sqrt[4]{a})^3 = (a^3)^{\frac{1}{4}} = \sqrt[4]{a^3} $$ $$(\sqrt{x^{3}})^4 = [(x^{3})^{\frac{1}{2}}]^4 = x^{(3)(\frac{1}{2})(4)} = x^6$$ $$x \sqrt{x}=(x^1)(x^{\frac{1}{2}}) = x^{1+\frac{1}{2}} = x^{\frac{3}{2}}$$ $$ \sqrt[3]{8x^6} = (8x^6)^{\frac{1}{3}} = (8)^{\frac{1}{3}} (x^6)^{\frac{1}{3}} = (2^3)^{\frac{1}{3}} (x)^{\frac{6}{3}} = 2x^2 $$ $$ \sqrt[6]{x^4 y^3 z^2} =(x^4 y^3 z^2)^{\frac{1}{6}} = x^{\frac{4}{6}} y^{\frac{3}{6}} z^{\frac{2}{6}} = x^{\frac{2}{3}} y^{\frac{1}{2}} z^{\frac{1}{3}} $$
Common Mistakes
Finally, I list the most common mistakes that students make when they have to simplify algebraic expressions with exponents.
$$
\begin{array}{|c|c|}
\hline
\textbf{Mistake} & \textbf{Example} \\ \hline
(3a)^3 \neq 3a^3 & \text{Instead} \quad (3a)^3 = (3^3)(a^3) = 27a^3 \\ \hline
\sqrt{a^2 + b^2} \neq a + b & \sqrt{3^2 + 4^2} \neq 3 + 4 \quad \text{instead} \quad \sqrt{3^2+4^2}=\sqrt{25}=5 \\ \hline
(a + b)^n \neq a^n + b^n & (2 + 3)^2 = 5^2 = 25 \neq 2^2 + 3^2 = 13 \\ \hline
(-a)^2 \neq -(a^2) & (-3)^2 = 9 \neq -(3^2) = -9 \quad \text{instead} \quad (-a)^2 = a^2 \\ \hline
\end{array}
$$
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