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SAT Math : Pattern Behaviors in Exponents

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Example Question #1 : Pattern Behaviors In Exponents

A five-year bond is opened with $\$5000$ in it and an interest rate of 2.5 %, compounded annually. This account is allowed to compound for five years. Which of the following most closely approximates the total amount in the account after that period of time?

Possible Answers:

$\$5811$

$\$6143$

$\$5657$

$\$5518$

$\$5625$

Correct answer:

$\$5657$

Explanation:

Each year, you can calculate your interest by multiplying the principle ( $\$5000$ ) by 1.025 . For one year, this would be:

1.025*5000=5125

For two years, it would be:

5125*1.025 , which is the same as 1.025*1.025*5000

Therefore, you can solve for a five year period by doing:

1.025^5*5000

Using your calculator, you can expand the 1.025^5 into a series of multiplications. This gives you 5657.041064453125 , which is closest to $\$5657$ .

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Example Question #3 : Pattern Behaviors In Exponents

If a cash deposit account is opened with $\$7500$ for a three year period at 3.5 % interest compounded once annually, which of the following is closest to the positive difference between the interest accrued in the third year and the interest accrued in the second year?

Possible Answers:

$\$0$

$\$281.2$

$\$11.41$

$\$9.51$

$\$81.41$

Correct answer:

$\$9.51$

Explanation:

It is easiest to break this down into steps. For each year, you will multiply by 1.035 to calculate the new value. Therefore, let's make a chart:

After year 1: 7500*1.035=7762.5 ; Total interest: 262.5

After year 2: 7762.5*1.035=8034.1875 ; Let us round this to 8034.19 ; Total interest: 271.69

After year 3: 8034.19 * 1.035 = 8315.38665 ; Let us round this to 8315.39 ; Total interest: 281.2

Thus, the positive difference of the interest from the last period and the interest from the first period is: 281.2-271.69=9.51

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Example Question #1 : How To Find Compound Interest

Jack has $\$15$ , 000 to invest. If he invests two-thirds of it into a high-yield savings account with an annual interest rate of $8\%$ , compounded quarterly, and the other third in a regular savings account at $6\%$ simple interest, how much does Jack earn after one year?

Possible Answers:

$\$901.43$

$\$1051.32$

$\$1081.98$

$\$1124.32$

$\$1128.75$

Correct answer:

$\$1124.32$

Explanation:

First, break the problem into two segments: the amount Jack invests in the high-yield savings, and the amount Jack invests in the simple interest account (10,000 and 5,000 respectively).

Now let's work with the high-yield savings account. $10,000 is invested at an annual rate of 8%, compounded quarterly. We can use the compound interest formula to solve:

$\text{Final Balance}=\text{ Principal} \cdot \bigg(1+\frac{\text{Interest Rate}}{c}\bigg)^{(\text{Time})(c)}$

Plug in the values given:

$=10,000\cdot \bigg(1+\frac{.08}{4}\bigg)^{(1)(4)}$

$=10,000 \cdot (1.02)^{4}$

=10,824.32

Therefore, Jack makes $824.32 off his high-yield savings account. Now let's calculate the other interest:

$\text{Interest}=\text{Principal} \cdot \text{ Interest Rate} \cdot \text{ Time}$

$=5,000 \cdot (0.06)\cdot (1)$

=300

Add the two together, and we see that Jack makes a total of, $\$1124.32$ off of his investments.

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Example Question #1 : How To Find Compound Interest

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= 6664.23

y= 5664.23

y= 6764.23

y= 6564.23

y= 16564

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 #251 : 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 #1 : Pattern Behaviors 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 #1 : How To Find Patterns In Exponents

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 #255 : 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 #256 : Exponents

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

Possible Answers:

1.079

1.116

1.592

1.346

1.255

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 #1 : How To Find Patterns In Exponents

Evaluate:

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

Possible Answers:

$x^{3}$

$x^{-3}$

$x^{-18}$

$x^{6}$

$x^{9}$

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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SAT Math : Pattern Behaviors in Exponents

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Pattern Behaviors In Exponents

Example Question #3 : Pattern Behaviors In Exponents

Example Question #1 : How To Find Compound Interest

Example Question #1 : How To Find Compound Interest

Example Question #251 : Exponents

Example Question #1 : Pattern Behaviors In Exponents

Example Question #1 : How To Find Patterns In Exponents

Example Question #255 : Exponents

Example Question #256 : Exponents

Example Question #1 : How To Find Patterns In Exponents

All SAT Math Resources

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