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ACT Math : Exponents

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Example Questions

Example Question #41 : Exponents

Simplify: hⁿ + h^–2n

Possible Answers:

^{$\displaystyle 2h^{-n}$}

^{$\displaystyle h^n+\frac{1}{h^{2n}}$}

$\displaystyle h$

^{$\displaystyle \frac{h^n}{h^{2n}}$}

$h^{-n}$ $\displaystyle h^{-n}$

Correct answer:

^{$\displaystyle h^n+\frac{1}{h^{2n}}$}

Explanation:

h^–2n= 1/h²ⁿ

hⁿ + h^–2n= hⁿ + 1/h²ⁿ

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Example Question #42 : Exponents

Simplify: 3y² + 7y² + 9y³ – y³ + y

Possible Answers:

10y⁴ + 8y⁶ + y

10y² + 10y³ + y

10y² + 8y³ + y

10y² + 9y³

19y¹¹

Correct answer:

10y² + 8y³ + y

Explanation:

Add the coefficients of similar variables (y, y², 9y³)

3y² + 7y² + 9y³ – y³ + y =

(3 + 7)y² + (9 – 1)y³ + y =

10y² + 8y³ + y

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Example Question #43 : Exponents

Simplify the following:

$\frac{K^3K^4}{M^6M^2}$ $\displaystyle \frac{K^3K^4}{M^6M^2}$

Possible Answers:

$\frac{K^{-1}}{M^4}$ $\displaystyle \frac{K^{-1}}{M^4}$

$\frac{KM^{\frac{1}{2}}}{KM^{\frac{1}{2}}}$ $\displaystyle \frac{KM^{\frac{1}{2}}}{KM^{\frac{1}{2}}}$

$\frac{K^8}{M^7}$ $\displaystyle \frac{K^8}{M^7}$

$\frac{K^7}{M^8}$ $\displaystyle \frac{K^7}{M^8}$

$\frac{K^{12}}{M^{12}}$ $\displaystyle \frac{K^{12}}{M^{12}}$

Correct answer:

$\frac{K^7}{M^8}$ $\displaystyle \frac{K^7}{M^8}$

Explanation:

When common variables have exponents that are multiplied, their exponents are added. So K³* K⁴ =K⁽³⁺⁴⁾ = K⁷. And M⁶ * M² = M⁽⁶⁺²⁾ = M⁸. So the answer is K⁷/M⁸.

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Example Question #44 : Exponents

Solve for $\displaystyle x$ :

$\frac{64^4}{8^x} = 8$ $\displaystyle \frac{64^4}{8^x} = 8$

Possible Answers:

x=-7 $\displaystyle x=-7$

x=7 $\displaystyle x=7$

x=1 $\displaystyle x=1$

x=0 $\displaystyle x=0$

x=8 $\displaystyle x=8$

Correct answer:

x=7 $\displaystyle x=7$

Explanation:

First, reduce all values to a common base using properties of exponents.

$\displaystyle 64^4 = (8^2)^4 = 8^8$

Plugging back into the equation-

$8^8 = 8^x\cdot 8^1$ $\displaystyle 8^8 = 8^x\cdot 8^1$

Using the formula

$x^a\cdot x^b=x ^{a+b}$ $\displaystyle x^a\cdot x^b=x ^{a+b}$

We can reduce our equation to

$\displaystyle x +1 = 8$

So,

x = 7 $\displaystyle x = 7$

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Example Question #45 : Exponents

Simplify: y³x⁴(yx³ + y²x² + y¹⁵ + x²²)

Possible Answers:

y⁴x⁷ + y⁵x⁶ + y¹⁸x⁴ + y³x²⁶

y³x¹² + y⁶x⁸ + y⁴⁵ + x⁸⁸

y³x¹² + y⁶x⁸ + y⁴⁵x⁴ + y³x⁸⁸

y³x¹² + y¹²x⁸ + y²⁴x⁴ + y³x²³

2x⁴y⁴ + 7y¹⁵ + 7x²²

Correct answer:

y⁴x⁷ + y⁵x⁶ + y¹⁸x⁴ + y³x²⁶

Explanation:

When you multiply exponents, you add the common bases:

y⁴ x⁷ + y⁵x⁶ + y¹⁸x⁴ + y³x²⁶

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Example Question #1 : Exponential Operations

If $\frac{3^{y - 1}}{3^{-2}} = 27^{y}3^{y}$ $\displaystyle \frac{3^{y - 1}}{3^{-2}} = 27^{y}3^{y}$ , what is the value of $\displaystyle y$ ?

Possible Answers:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

$\displaystyle 3$

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

$\displaystyle 4$

Correct answer:

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

Explanation:

Rewrite the term on the left as a product. Remember that negative exponents shift their position in a fraction (denominator to numerator).

$3^{y-1}*3^2=27^y3^y$ $\displaystyle 3^{y-1}*3^2=27^y3^y$

The term on the right can be rewritten, as 27 is equal to 3 to the third power.

$3^{y-1}*3^2=(3^3)^y*3^y$ $\displaystyle 3^{y-1}*3^2=(3^3)^y*3^y$

Exponent rules dictate that multiplying terms allows us to add their exponents, while one term raised to another allows us to multiply exponents.

$3^{(y-1)+2}=(3)^{3y}*3^y$ $\displaystyle 3^{(y-1)+2}=(3)^{3y}*3^y$

$3^{y+1}=3^{3y+y}=3^{4y}$ $\displaystyle 3^{y+1}=3^{3y+y}=3^{4y}$

We now know that the exponents must be equal, and can solve for $\displaystyle y$ .

y+1=4y $\displaystyle y+1=4y$

1=3y $\displaystyle 1=3y$

$\frac{1}{3}=y$ $\displaystyle \frac{1}{3}=y$

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Example Question #2 : How To Add Exponents

If $5^2 \times 5^n = 5^{12}$ $\displaystyle 5^2 \times 5^n = 5^{12}$ , what is the value of $\displaystyle n$ ?

Possible Answers:

$\displaystyle 4$

$\displaystyle 10$

$\displaystyle 24$

$\displaystyle 14$

$\displaystyle 6$

Correct answer:

$\displaystyle 10$

Explanation:

Since the base is 5 for each term, we can say 2 + n =12. Solve the equation for n by subtracting 2 from both sides to get n = 10.

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Example Question #2233 : Act Math

Which expression is equivalent to the following? $\displaystyle 2^2 + 2^2 + 2^3$

Possible Answers:

$2^{12}$ $\displaystyle 2^{12}$

None of these

2^7 $\displaystyle 2^7$

$\displaystyle 4^2 + 2^3$

6^7 $\displaystyle 6^7$

Correct answer:

None of these

Explanation:

The rule for adding exponents is $a^m \cdot a^n = a^{m+n}$ $\displaystyle a^m \cdot a^n = a^{m+n}$ . We can thus see that 2^2 $\displaystyle 2^2$ and 2^3 $\displaystyle 2^3$ are no more compatible for addition than x^2 $\displaystyle x^2$ and x^3 $\displaystyle x^3$ are.

You could combine the first two terms into 2(2^2) $\displaystyle 2(2^2)$ , but note that PEMDAS prevents us from equating this to 4^2 $\displaystyle 4^2$ (the exponent must solve before the distribution).

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Example Question #2234 : Act Math

Express as a power of 2: $2(4^3) \cdot 4(2^3)$ $\displaystyle 2(4^3) \cdot 4(2^3)$

Possible Answers:

$2^{12}$ $\displaystyle 2^{12}$

$2^{16}$ $\displaystyle 2^{16}$

The expression cannot be rephrased as a power of 2.

2^7 $\displaystyle 2^7$

$2^{22}$ $\displaystyle 2^{22}$

Correct answer:

$2^{12}$ $\displaystyle 2^{12}$

Explanation:

Since the problem requires us to finish in a power of 2, it's easiest to begin by reducing all terms to powers of 2. Fortunately, we do not need to use logarithms to do so here.

$\displaystyle 2(4^3) = 2^1((2^2)^3)$

$\displaystyle 4(2^3) = 2^2(2^3)$

Thus, $2(4^3) \cdot 4(2^3) = 2^1((2^2)^3) \cdot 2^2(2^3) = 2^1(2^6) \cdot (2^5) = 2^7 \cdot 2^5 = 2^{12}$ $\displaystyle 2(4^3) \cdot 4(2^3) = 2^1((2^2)^3) \cdot 2^2(2^3) = 2^1(2^6) \cdot (2^5) = 2^7 \cdot 2^5 = 2^{12}$

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Example Question #46 : Exponents

Simplify the following expression:
$\displaystyle x^3*x^5*y^2*y^7$

Possible Answers:

$x*y^{17}$ $\displaystyle x*y^{17}$

$(x*y)^{17}$ $\displaystyle (x*y)^{17}$

$(x*y)^{210}$ $\displaystyle (x*y)^{210}$

x^8*y^9 $\displaystyle x^8*y^9$

$x^{15}*y^{14}$ $\displaystyle x^{15}*y^{14}$

Correct answer:

x^8*y^9 $\displaystyle x^8*y^9$

Explanation:

When multiplying bases that have exponents, simply add the exponents. Note that you can only add the exponents if the bases are the same. Thus:

$\displaystyle x^3*x^5*y^2*y^7$ $= x^{3+5}*y^{2+7} = x^8*y^9$ $\displaystyle = x^{3+5}*y^{2+7} = x^8*y^9$

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ACT Math : Exponents

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #41 : Exponents

Example Question #42 : Exponents

Example Question #43 : Exponents

Example Question #44 : Exponents

Example Question #45 : Exponents

Example Question #1 : Exponential Operations

Example Question #2 : How To Add Exponents

Example Question #2233 : Act Math

Example Question #2234 : Act Math

Example Question #46 : Exponents

All ACT Math Resources

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