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SAT Math : Exponents

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Example Questions

Example Question #2431 : Sat Mathematics

A truck was bought for $\$ 20,000$ in 2008, and it depreciates at a rate of $13\%$ per year. What is the value of the truck in 2016? Round to the nearest cent.

Possible Answers:

y= 6764.23

y= 16564

y= 6664.23

y= 6564.23

y= 5664.23

Correct answer:

y= 6564.23

Explanation:

The first step is to convert the depreciation rate into a decimal. $13\%=0.13$ . Now lets recall the exponential decay model. y=C(1-r)^t , where is the starting amount of money, is the annual rate of decay, and is time (in years). After substituting, we get

y=20000(1-0.13)^8

y=20000(0.87)^8

$y=20000\cdot 0.3282117$

y= 6564.23

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Example Question #1 : How To Find Patterns In Exponents

If a^x·a⁴ = a¹² and (b^y)³ = b¹⁵, what is the value of x - y?

Possible Answers:

-4

-2

-9

Correct answer:

Explanation:

Multiplying like bases means add the exponents, so x+4 = 12, or x = 8.

Raising a power to a power means multiply the exponents, so 3y = 15, or y = 5.

x - y = 8 - 5 = 3.

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Example Question #2 : How To Find Patterns In Exponents

If p and q are positive integrers and 27^p= 9^q, then what is the value of q in terms of p?

Possible Answers:

(2/3)p

(3/2)p

Correct answer:

(3/2)p

Explanation:

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 3^3p= 3^2q. So then 3p = 2q, and q = (3/2)p is our answer.

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Example Question #2432 : Sat Mathematics

Simplify 27^2/3.

Possible Answers:

125

729

Correct answer:

Explanation:

27^2/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations.

27^2/3 = (27²)^1/3 = 729^1/3 OR

27^2/3 = (27^1/3)² = 3²

Obviously 3² is much easier. Either 3² or 729^1/3 will give us the correct answer of 9, but with 3² it is readily apparent.

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Example Question #3 : How To Find Patterns In Exponents

If and are integers and

$\left ( \frac{1}{3} \right )^{a}=27^{b}$

what is the value of $a\div b$ ?

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

Correct answer:

Explanation:

To solve this problem, we will have to take the log of both sides to bring down our exponents. By doing this, we will get $\dpi{100} \small a\ast log\left (\frac{1}{3} \right )= b\ast log\left ( 27 \right )$ .

To solve for $\dpi{100} \small \frac{a}{b}$ we will have to divide both sides of our equation by $\dpi{100} \small log\frac{1}{3}$ to get $\dpi{100} \small \frac{a}{b}=\frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}$ .

$\dpi{100} \small \frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}$ will give you the answer of –3.

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Example Question #4 : How To Find Patterns In Exponents

If $\log 2=0.301$ and $\log 3=0.477$ , then what is $\log 12$ ?

Possible Answers:

1.592

1.255

1.116

1.346

1.079

Correct answer:

1.079

Explanation:

We use two properties of logarithms:

$log(xy) = log (x) + log (y)$

$log(x^{n}) = nlog (x)$

So $\log 12=2 \log2+\log3$

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Example Question #10 : Pattern Behaviors In Exponents

Evaluate:

$x^{-3}x^{6}$

Possible Answers:

$x^{-3}$

$x^{6}$

$x^{9}$

$x^{-18}$

$x^{3}$

Correct answer:

$x^{3}$

Explanation:

$x^{m}\ast x^{n} = x^{m + n}$ , here m=-3 and n=6 , hence -3+6=3 .

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Example Question #254 : Exponents

Solve for

$\left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}$

Possible Answers:

None of the above

Correct answer:

Explanation:

$\left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}$ = $\left ( \frac{3}{2} \right )^{3} = \left ( \frac{2}{3} \right )^{-3}$

x+1=-3 which means x=-4

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Example Question #255 : Exponents

Which of the following statements is the same as:

$\dfrac{4^{x+2}}{2^y}$

$I)\ 2^{2x+4-y} \quad II)\ 2^{x+2} / 4^y \quad III)\ (4^x)( 4^2 )(2^{-y} )$

Possible Answers:

$II \mbox{ and } III \mbox{ only}$

$I \mbox{ and } III \mbox{ only}$

$I, II \mbox{ and } III$

$I \mbox{ and } II \mbox{ only}$

$I \mbox{ only}$

Correct answer:

$I \mbox{ and } III \mbox{ only}$

Explanation:

Remember the laws of exponents. In particular, when the base is nonzero:

$b^{m+n} = b^m b^n \qquad (b^m)^n = b^{mn} \\ (b \times c)^n = b^n c^n \qquad b^{-n} = 1/b^n$

An effective way to compare these statements, is to convert them all into exponents with base 2. The original statement becomes:

$4^{x+2} / 2^y = 4^{x+2} (2^{-y}) = (2^2)^{x+2} 2^{-y} = 2^{2(x+2) } 2^{-y} = 2^{2x+4-y}$

This is identical to statement I. Now consider statement II:

$2^{x+2} / 4^y = 2^{x+2} ( 4^{-y} ) = 2^{x+2} ( 2^2 )^{-y} = 2^{x+2} 2^{-2y} = 2^{x+2-2y} \neq 2^{2x+4-y}$

Therefore, statement II is not identical to the original statement. Finally, consider statement III:

$(4^x)(4^2 ) (2^{-y}) = (2^2)^x (2^2)^2 (2^{-y}) = (2^{2x} )( 2^4 ) (2^{-y}) = 2^{2x+4-y}$

which is also identical to the original statement. As a result, only I and III are the same as the original statement.

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Example Question #256 : Exponents

Write in exponential form:

$\sqrt[3]{\left ( x-1 \right )^{2}}$

Possible Answers:

$\left ( x-1 \right )$

$\left ( x-1 \right )^{3}$

$\left ( x-1 \right )^{2}$

$\left ( x-1 \right )^{\frac{2}{3}}$

$\left ( x-1 \right )^{\frac{3}{2}}$

Correct answer:

$\left ( x-1 \right )^{\frac{2}{3}}$

Explanation:

Using properties of radicals e.g., $\sqrt[m]{x^{n}} = x^{\frac{n}{m}}$

we get $\sqrt[3]{\left ( x-1 \right )^{2}} = \left ( x-1 \right )^{\frac{2}{3}}$

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SAT Math : Exponents

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2431 : Sat Mathematics

Example Question #1 : How To Find Patterns In Exponents

Example Question #2 : How To Find Patterns In Exponents

Example Question #2432 : Sat Mathematics

Example Question #3 : How To Find Patterns In Exponents

Example Question #4 : How To Find Patterns In Exponents

Example Question #10 : Pattern Behaviors In Exponents

Example Question #254 : Exponents

Example Question #255 : Exponents

Example Question #256 : Exponents

All SAT Math Resources

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