Pattern Behaviors in Exponents - PSAT Math

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Question

Donna wants to deposit money into a certificate of deposit so that in exactly ten years, her investment will be worth \$100,000. The interest rate of the CD is 7.885%, compounded monthly.

What should Donna's initial investment be, at minimum?

Answer

The formula for compound interest is

$A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}$

where $A_{0}$ is the initial investment, is the interest rate expressed as the decimal equivalent, is the number of periods per year the interest is compounded, is the number of years, and is the final value of the investment.

Set (monthly = 12 periods), and , and evaluate $A_{0}$ :

$100,000 = A_{0} \left ( 1+ \frac{0.07885}{12} \right )^{12 \cdot 10}$

$100,000 \approx A_{0} \left ( 1+0.0065708\right )^{120}$

$100,000 \approx A_{0} \left ( 1 .0065708\right )^{120}$

$100,000 \approx A_{0} \cdot 2.1944$

$A_{0} \approx 100,000 \div 2.1944$

$A_{0} \approx 45,569.99$

The correct response is \$45,569.99.

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Question

On January 15, 2015, Philip deposited \$10,000 in a certificate of deposit that returned interest at an annual rate of 8.125%, compounded monthly. How much will his certificate of deposit be worth on January 15, 2020?

Answer

The formula for compound interest is

$A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}$

where $A_{0}$ is the initial investment, is the interest rate expressed as the decimal equivalent, is the number of periods per year the interest is compounded, is the number of years, and is the final value of the investment.

Set $A_{0} = 10,000, r = 0.08125, N = 12$ (monthly = 12 periods), and , and evaluate :

$A = 10,000 \left ( 1+ \frac{0.08125}{12} \right )^{12 \cdot 5}$

$A \approx 10,000 \left ( 1.0067708\right )^{60}$

$A \approx 10,000 \cdot 1.499124$

$A \approx 1 4,991.24$

The CD will be worth \$14,991.24.

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Question

Money is deposited in corporate bonds which yield 6.735% annual interest compounded monthly, and which mature after ten years. Which of the following responses comes closest to the percent by which the value of bonds increases?

Answer

The formula for compound interest is

$A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}$

where $A_{0}$ is the initial investment, is the interest rate expressed as the decimal equivalent, is the number of periods per year the interest is compounded, is the number of years, and is the final value of the investment.

In the given scenario, , , and (monthly); substitute:

$A = A_{0}\left ( 1+ \frac{0.06735}{12} \right )^{12 \cdot 10}$

$A \approx A_{0}\left ( 1+0.0056125 \right )^{12 0}$

$A \approx A_{0}\left ( 1 .0056125 \right )^{12 0}$

$A \approx A_{0} \cdot 1.9574$

This meas that the final value of the bonds is about 1.96 times their initial value, or, equivalently, 96% greater than their initial value. Of the given responses, 95% comes closest.

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Question

Tom invests $\$15$ , in a savings account with an annual interest rate of . If his investment is compounded semiannually, how much interest does he earn after years?

Answer

In order to find the interest earned, used the compound interest formula

$\text{Final Balance} = \text{Principal} \cdot (1+\frac{IR}{C})^{(\text{Time})(\text{C})}$

where represents the number of times the account is compounded each year, and represents the interest rate expressed as a decimal.

$=15,000\cdot(1+\frac{.06}{2})^{(2)(2)}$

$=15,000\cdot(1+.03)^{4}$

$=15,000\cdot(1.03)^{4}$

The account is worth \$16882.63 after two years. Therefore Tom earns \$1882.63 in interest.

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Question

If is the complex number such that , evaluate the following expression:

$i^{25}-i^{23}+i^{21}-i^{19}+i^{17}-i^{15}.\ .\ .+i$

Answer

The powers of i form a sequence that repeats every four terms.

i1 = i

i2 = -1

i3 = -i

i4 = 1

i5 = i

Thus:

i25 = i

i23 = -i

i21 = i

i19= -i

Now we can evalulate the expression.

i25 - i23 + i21 - i19 + i17..... + i

= i + (-1)(-i) + i + (-1)(i) ..... + i

= i + i + i + i + ..... + i

Each term reduces to +i. Since there are 13 terms in the expression, the final result is 13i.

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Question

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

Answer

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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Question

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

Answer

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 33p = 32q. So then 3p = 2q, and q = (3/2)p is our answer.

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Question

Simplify 272/3.

Answer

272/3 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.

272/3 = (272)1/3 = 7291/3 OR

272/3 = (271/3)2 = 32

Obviously 32 is much easier. Either 32 or 7291/3 will give us the correct answer of 9, but with 32 it is readily apparent.

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Question

If and are integers and

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

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

Answer

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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Question

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

Answer

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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Question

Evaluate:

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

Answer

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

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Question

Solve for

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

Answer

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

which means

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Question

Which of the following statements is the same as:

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

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

Answer

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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Question

Write in exponential form:

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

Answer

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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Question

Write in exponential form:

$\sqrt[3]{x^{5}}$

Answer

Properties of Radicals

$\sqrt[]{x} = x^{\frac{1}{2}}$

$\sqrt[3]{x} = x^{\frac{1}{3}}$

$x^{\frac{5}{3}}$

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Question

Write in radical notation:

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

Answer

Properties of Radicals

$\sqrt[]{x} = x^{\frac{1}{2}}$

$\sqrt[3]{x} = x^{\frac{1}{3}}$

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Question

Express in radical form : $x^{\frac{3}{5}}$

Answer

Properties of Radicals

$\sqrt[]{x} = x^{\frac{1}{2}}$

$\sqrt[3]{x} = x^{\frac{1}{3}}$

$x^{\frac{3}{5}}=\sqrt[5]{x^{3}}$

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Question

Simplify: $\sqrt[3]{2}\star \sqrt[3]{32}$

Answer

$\sqrt[3]{2}\star \sqrt[3]{32}= \sqrt[3]{64}= 4$

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Question

Simplify:

$\frac{\sqrt[3]{x^{5}y^{-2}}}{\sqrt[3]{x^{-1}y^{4}}}$

Answer

Convert the given expression into a single radical e.g. $\sqrt[3]{\frac{x^{5}y^{-2}}{x^{-1}y^{4}}}$ the expression inside the radical is:

$\frac{x^{6}}{y^{6}}$

and the cube root of this is :

$\frac{x^{2}}{y^{2}}$

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Question

Solve for .

$\log _{\frac{1}{5}}\left ( \frac{1}{25} \right )=x$

Answer

$\frac{1}{25}= \left ( \frac{1}{5} \right )^{x}$

Hence must be equal to 2.

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