[fs-toc-h2]What are Exponents?

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Exponents are a shorthand for multiplication, the same way multiplication is a shorthand for addition.

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\[2 \times 2 \times 2 \times 2 = 2^4\]

\[3 \times 3 \times 3 = 3^3\]

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[fs-toc-h2]Exponent Rules (IMPORTANT)

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When asked a question about exponents on the SAT, 95% of the time it will either be testing exponent rules or exponential relationships.

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Many students freeze up when they see these questions, as three different variables all attached with exponents may seem a bit intimidating. Luckily though, every single exponent rule question can be answered (fairly easily too) with knowledge of X exponent rules.

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Note: \(x = x^1\)

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[fs-toc-h3]Rule 1

\(\boldsymbol{a^m \times a^n = a^{m+n}}\)

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In words, when you multiply two terms with the same base, you add their exponents. For example,

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\[a^2 \times a^4 = a^{2+4} = a^6\]

\[3^2 \times 3^3 = 3^{2+3} = 3^5\]

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This rule still holds up when a coefficient is added in front of the variable. As long as they are ‘like terms,’ you can multiply the bases and add the exponents.

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\[3x \times 4x^3 = ?\]

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First, multiply the bases: \(4x \times 3x = 12x\). Then, add the exponents: \(1+3 = 4\). So, our final answer is \(12x^4\).

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This may seem a bit complex when multiple variables are added to the problem, though when taking it one step at a time, it's just as easy. For example:

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\[(x^3)(x^4 + a^7)\]

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Remember, you can only simplify exponents when the bases are like terms. In this case, we have two variables, \(x\) and \(a\). First, ignore all of the ‘a’ terms and begin to simplify all the ‘x’ terms using the distributive property:

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\[(x^3)(x^4) = x^{3+4} = x^7\]

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Now, distribute the \(x\) term to the \(a\) term. Remember though, you cannot add their exponents, as they are different bases, so you're left with:

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\[(x^3)(a^7) = x^3a^7\]

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Adding our two expressions together, the final answer becomes:

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\[x^7 + x^3a^7\]

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[fs-toc-h3]Rule 2

\(\boldsymbol{\frac{a^m}{a^n} = a^{m-n}}\)

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When you divide two terms of the same base, you subtract their exponents. For example,

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\[\frac{x^4}{x^2} = x^{4-2} = x^2\]

\[\frac{6x^5}{2x^3} = 3x^{5-3} = 3x^2\]

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Just like multiplying, you can only simplify and subtract exponents if the bases are like terms.

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\[\frac{x^6y^4}{x^2y^2} = x^{6-2}y^{4-2} = x^4y^2\]

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[fs-toc-h3]Rule 3

\(\boldsymbol{(a^m)^n = a^{m \times n}}\)

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When raising an exponential term to another power, the exponents are multiplied.

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\[(3^2)^4 = 3^{2 \times 4} = 3^8\]

\[(x^3y^2)^4 = x^{3 \times 4}y^{2 \times 4} = x^{12}y^8\]

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This is most likely the easiest exponent rule to use since you don't have to worry about the base; you simply distribute the exponent to each term under it.

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[fs-toc-h3]Rule 4

\(\boldsymbol{\sqrt[n]{a^m} = a^{m/n}}\)

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Roots can be written as exponents, and vice versa. Though, if asked to simplify an expression on the SAT, you'll most likely be asked to convert a root into an exponent, but the other way around has been tested before.

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\[\sqrt{a} = a^{1/2}\]

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Note: A root symbol without a number on the outside \((\sqrt{x})\) is the same as \(x^{1/2}\), known as a square root.

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\[\sqrt[4]{a^8} = a^{8/4} = a^2\]

\[\sqrt[3]{x^2y^3} = x^{2/3}y^{3/3} = x^{2/3}y\]

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Note: Again, anything raised to the power of one is just left blank; \(y^1 = y\)

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[fs-toc-h3]Rule 5

\(\boldsymbol{a^{-m} = \frac{1}{a^m}}\)

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Whenever a term is raised to a negative power, it can be written as a fractional expression.

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\[a^{-3} = \frac{1}{a^3}\]

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When you have a negative exponent that's already in a fraction, simply switch its position: if it's in the numerator, send it down to the denominator. If it's in the denominator, send it up to the numerator.

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\[\frac{a^{-2}}{x^{-4}} = \frac{x^4}{a^2}\]

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If negative and positive exponential terms are being multiplied, you can simply ‘detach’ them from each other:

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\[\frac{x^2 y^{-3}}{z^{-5}} = \frac{x^2 z^5}{y^3}\]

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[fs-toc-h3]Rule 6

\(\boldsymbol{\left(\frac{a}{x}\right)^m = \frac{a^m}{x^m}}\)

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When a fraction is raised to a power, the exponent distributes to each term in the numerator and denominator.

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\[\left(\frac{x}{y}\right)^2 = \frac{x^2}{y^2}\]

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The same idea holds up when multiple terms are added into both the numerator and denominator.

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\[\left(\frac{xy}{ab}\right)^3 = \frac{x^3y^3}{a^3b^3}\]

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If the term inside the fraction already has an exponent, multiply it by the other power it’s raised to, as seen in rule 3:

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\[\left(\frac{z^3x^2y^4}{p^2}\right)^2 = \frac{z^{3 \times 2}x^{2 \times 2}y^{4 \times 2}}{p^{2 \times 2}} = \frac{z^6x^4y^8}{p^4}\]

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[fs-toc-h3]Rule 7

\(\boldsymbol{a^0 = 1}\)

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ANYTHING to the power of zero is one, no matter if it is a variable or number, it will always be 1.

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\[x^0 = 1\]

\[3^0 = 1\]

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[fs-toc-h2]Combining Rules

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Sometimes, the SAT will test you on only one rule, other times they’ll test you on multiple. These questions are almost always the most intimidating as they can involve 4+ variables, multiple roots, a fraction, and exponents. Though, when taken one step at a time, simplifying using 4 rules is the exact same as using 1. Let's look at some SAT questions that you may be asked.

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[fs-toc-h4]Example 1

This first one is simpler, as it DOES only test one rule, though they do add an extra step as you may quickly realize.

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Rule 4 says that \(a^{11/12}\) is equal to \(\sqrt[12]{a^{11}}\). However, when looking at the answer choices, this doesn’t show up at all.

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Note: If the ratio between \(m\) and \(n\) (in \(a^{m/n}\)) is equivalent, then the whole expression is equivalent. For example:

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\[a^{4/2} = a^{10/5}\]

\[ \text{Or}\]

\[2a^4 = 5a^{10}\]

\[\sqrt[2]{a^4} = \sqrt[5]{a^{10}}\]

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Since \(4/2\) and \(10/5\) both equal 2, then we can say that both of these expressions are equivalent.

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So back to the question, to find which expression is equivalent, we have to find which two numbers are equal to the ratio \(11/12\). Simply go down the line, plug the fractions into your calculator, and see which ratio is correct.

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Choice A: Here the two numbers are 132 and 12. \(\frac{132}{12}\) is NOT equivalent to \(\frac{11}{12}\), meaning we can throw it out.

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Choice B: Here we have \(\frac{132}{144}\). When put into a calculator, we see that this does in fact simplify to \(\frac{11}{12}\). \(\left(\frac{132}{144} = \frac{11}{12}\right)\). This means that the expression is equivalent to our original one, meaning that choice B is the correct answer.

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[fs-toc-h4]Example 2

This question is also quite straightforward, as it only tests one rule: \((a^m \times a^n = a^{m+n})\).

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Remember, you can only simplify exponents when they have the same base. So, multiply the \(m\) terms, the \(q\) terms, then the \(z\) terms all by themselves.

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First, multiply the two \(m\) terms:

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\[m^4 \times m^1 = m^{4+1} = m^5\]

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Then the \(q\) terms:

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\[q^4 \times q^5 = q^{4+5} = q^9\]

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Finally, the \(z\) terms:

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\[z^{-1} \times z^3 = z^{2}\]

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Putting all terms together again, the final answer is \(m^5q^9z^2\), which corresponds to answer choice B.

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Tip: When simplifying the \(m\) term and getting \(m^5\), this eliminated answer choices A, C, and D since all of their \(m\) terms have different exponents, meaning you didn't actually have to simplify the whole expression.

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This question is testing us on multiple rules. The question first wants us to simplify this long expression into the \(ax^b\) form.

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While this may look a bit intimidating, take it step by step. Firstly, get rid of all the radicals and transform them into exponents.

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\[6 \cdot 3^{5/5} \cdot x^{45/5} \cdot 2^{8/8} \cdot x^{1/8}\]

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Simplify all fractions in the exponents:

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\[6 \cdot 3 \cdot x^9 \cdot 2 \cdot x^{1/8}\]

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Now, multiply all numbers and variables together to fully simplify:

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\[18x^9 \cdot 2x^{1/8}\]

\[36x^{9+1/8}\]

\[36x^{73/8}\]

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Our simplified version of the expression is \(36x^{73/8}\), but the question isn’t just asking for an expression, it's asking for the value of \(a + b\).

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In this case, \(36 = a\) and \(73/8 = b\).

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\[36 + \frac{73}{8} = \frac{361}{8}\]

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\(\frac{361}{8}\) is the final answer.