Exponents - SAT Math

Card 0 of 2288

Question

If and are positive integers and , then what is the value of $\frac{m}{n}$ ?

Answer

43 = 64

Alternatively written, this is 4(4)(4) = 64 or 43 = 641.

Thus, m = 3 and n = 1.

m/n = 3/1 = 3.

Compare your answer with the correct one above

Question

Write the following logarithm in expanded form:

$\log x^{2}y$

Answer

$\log x^{2}y=\log x^{2}+\log y=2\log x+\log y$

Compare your answer with the correct one above

Question

If and are both rational numbers and $4^{m} = 8^{n}$ , what is $\frac{m}{n}$ ?

Answer

This question is asking you for the ratio of m to n. To figure it out, the easiest way is to figure out when 4 to an exponent equals 8 to an exponent. The easiest way to do that is to list the first few results of 4 to an exponent and 8 to an exponent and check to see if any match up, before resorting to more drastic means of finding a formula.

$4^{1} = 4, 4^{2} = 16, 4^{3} = 64, 4^{4} = 256, 4^{5} = 1024$

$8^{1}=8, 8^{2} = 64, 8^{3} = 512, 8^{4} = 4096, 8^{5} = 32768$

And, would you look at that. $4^{3}= 8^{2}$ . Therefore, $\frac{m}{n} = \frac{3}{2}$ .

Compare your answer with the correct one above

Question

Let . What is the following equivalent to, in terms of :

$\frac{x}{\sqrt{y}}$

Answer

Solve for x first in terms of y, and plug back into the equation.

$\4x^2 = y^3 \ \x^2 = \frac{1}{4} y^3 \ \x = \pm \frac{1}{2} y\sqrt{y}$

Then go back to the equation you are solving for:

$\frac{x}{\sqrt{y}}$ substitute in $x=\pm\frac{1}{2}y\sqrt{y}$

$\ \frac{x}{\sqrt{y}}=\frac{\pm\frac{1}{2}y\sqrt{y}}{\sqrt{y}} \ \\frac{x}{\sqrt{y}}=\pm\frac{1}{2}y$

Compare your answer with the correct one above

Question

For which of the following values of is the value of ${x^{-\frac{1}{2}}}$ least?

Answer

${x^{-\frac{1}{2}}}$ is the same as $\frac{1}{\sqrt{x}}$ , which means that the bigger the answer to $\sqrt{x}$ is, the smaller the fraction will be.

Therefore, is the correct answer because

$\frac{1}{\sqrt{16}} = \frac{1}{4}$ .

Compare your answer with the correct one above

Question

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(2+i) \ast i$ .

Answer

$a \ast b = \frac{a + b}{a}$ , so

$(2+i) \ast i = \frac{(2+ i )+ i}{2+i} = \frac{2 + 2i}{2+i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$(2+i) \ast i$

$= \frac{2 + 2i}{2+i}$

$= \frac{\left (2 + 2i \right )\left (2-i \right )}{\left (2+i \right )\left (2-i \right )}$

$= \frac{ 2 \cdot 2 - 2 \cdot i + 2i \cdot 2 - 2i \cdot i}{2 ^{2}+1^{2}}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot i^{2}}{4+1}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot (-1)}{5}$

$= \frac{ 4 - 2 i + 4i +2}{5}$

$= \frac{ 6 + 2 i }{5}$

$= \frac{ 6 }{5} + \frac{ 2 }{5}i$

Compare your answer with the correct one above

Question

Simplify the expression by rationalizing the denominator, and write the result in standard form: $\frac{5+i}{7-i}$

$\left (\textup{Note: } i=\sqrt{-1}\right\ )$

Answer

Multiply both numerator and denominator by the complex conjugate of the denominator, which is :

$\frac{5+i}{7-i}$

$= \frac{(5+i)(7+i)}{(7-i)(7+i)}$

$= \frac{5 \cdot 7 + 5 \cdot i + 7 \cdot i + i \cdot i }{(7^{2}+1 ^{2}) }$

$= \frac{35 +5i + 7 i +(-1) }{(49+1) }$

$= \frac{34 +12i }{50 }$

$= \frac{34 }{50 } + \frac{ 12i }{50 }$

$= \frac{17}{25} + \frac{ 6}{25}i$

Compare your answer with the correct one above

Question

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(3+2i) \ast 2$

Answer

$a \ast b = \frac{a + b}{a}$ , so

$(3+2i) \ast 2 = \frac{(3+2i) + 2}{3+2i } = \frac{5+2i}{3+2i }$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$(3+2i) \ast 2 = \frac{5+2i}{3+2i }$

$= \frac{\left (5+2i \right )\left (3-2i \right )}{\left (3+2i \right )\left (3-2i \right )}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2i \cdot 2i }{3^{2}+2^{2}}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2 \cdot 2 \cdot i^{2} }{9+4}$

$= \frac{15 - 10i + 6i -4 \cdot (-1) }{9+4}$

$= \frac{15 - 10i + 6i +4}{13}$

$= \frac{19 -4i}{13}$

$= \frac{19 }{13} - \frac{ 4}{13}i$

Compare your answer with the correct one above

Question

Define an operation $\Phi$ such that, for any complex number ,

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

Answer

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

, and

$N=\frac{ i}{ 1+ 2i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is :

$N=\frac{ i}{ 1+ 2i}$

$=\frac{ i( 1- 2i)}{( 1+ 2i)( 1- 2i)}$

$=\frac{ i \cdot 1- i \cdot 2i}{1 ^{2}+2^{2}}$

$=\frac{ i - 2i^{2}}{1+4}$

$=\frac{ i - 2(-1)}{5}$

$=\frac{ i +2}{5}$

$=\frac{ 2}{5} + \frac{1}{5} i$

Compare your answer with the correct one above

Question

Define an operation $\Theta$ such that for any complex number ,

$\Theta a = 2a i + i$

If $\Theta N = 1,000$ , evaluate .

Answer

First substitute our variable N in where ever there is an a.

Thus, $\Theta a = 2a i + i$ , becomes $\Theta N = 2N i + i$ .

Since $\Theta N = 1,000$ , substitute:

In order to solve for the variable we will need to isolate the variable on one side with all other constants on the other side. To do this, apply the oppisite operation to the function.

First subtract i from both sides.

Now divide by 2i on both sides.

$\frac{2N i}{2i}= \frac{1,000 - i}{2i}$

From here multiply the numerator and denominator by i to further solve.

$N= \frac{1,000 - i}{2i}$

$= \frac{(1,000 - i) \cdot i }{2i \cdot i}$

$= \frac{1,000\cdot i - i\cdot i }{2 \cdot i^{2}}$

Recall that by definition. Therefore,

$= \frac{1,000 i - (-1) }{2 \cdot (-1)}$

$= \frac{1,000 i +1 }{-2}$

$= \frac{ 1 }{-2}+ \frac{1,000 i }{-2}$

$=- \frac{ 1 }{2}- 500i$ .

Compare your answer with the correct one above

Question

Define an operation $\blacklozenge$ as follows:

For any two complex numbers and ,

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$

Evaluate $(1+ i) \blacklozenge (1+ i)$ .

Answer

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$ , so

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

We can simplify each expression separately by rationalizing the denominators.

$\frac{1+ i}{2i}$

$= \frac{\left (1+ i \right )\cdot i}{2i \cdot i}$

$= \frac{1 \cdot i+ i \cdot i}{2 \cdot i ^{2}}$

$= \frac{ i+(-1)}{2 (-1)}$

$= \frac{-1+i}{-2}$

$= \frac{ 1 }{2} - \frac{1}{2} i$

$\frac{2i}{1+ i}$

$= \frac{2i(1- i)}{(1+ i)(1- i)}$

$= \frac{2i \cdot 1- 2i \cdot i}{1^{2}+1^{2}}$

$= \frac{2i - 2i ^{2}}{1+1}$

$= \frac{2i - 2(-1)}{2}$

$= \frac{2i +2}{2}$

Therefore,

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

$=\left ( \frac{ 1 }{2} - \frac{1}{2} i \right )+ \left (1 + i \right )$

$=\left ( \frac{ 1 }{2}+1 \right )+ \left ( i- \frac{1}{2} i \right )$

$= \frac{ 3 }{2} + \frac{1}{2} i$

Compare your answer with the correct one above

Question

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

Answer

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:

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

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

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

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

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

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

Compare your answer with the correct one above

Question

A five-year bond is opened with $\$5000$ in it and an interest rate of %, 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?

Answer

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

For two years, it would be:

, which is the same as

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

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

Compare your answer with the correct one above

Question

If a cash deposit account is opened with $\$7500$ for a three year period at % 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?

Answer

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

After year 1: ; Total interest:

After year 2: ; Let us round this to ; Total interest:

After year 3: ; Let us round this to ; Total interest:

Thus, the positive difference of the interest from the last period and the interest from the first period is:

Compare your answer with the correct one above

Question

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

Answer

The first step is to convert the depreciation rate into a decimal. . Now lets recall the exponential decay model. , where is the starting amount of money, is the annual rate of decay, and is time (in years). After substituting, we get

$y=20000\cdot 0.3282117$

Compare your answer with the correct one above

Question

$Simplify: \frac{\sqrt[3]{81}}{\sqrt[3]{3}}$

Answer

$\frac{\sqrt[3]{81}}{\sqrt[3]{3}} = \sqrt[3]{\frac{81}{3}}= \sqrt[3]{27}= 3$

Compare your answer with the correct one above

Question

$\sqrt{x+3} -x = 1$

Answer

From the equation in the problem statement

$\sqrt{x+3} = x + 1$

Now squaring both sides we get

$x+3=x^{2}+2x+1$ this is a quadratic equation which equals

$x^{2}+x -2 = 0$

and the factors of this equation are

$\left ( x+2 \right )\left ( x-1 \right )$

This gives us $x=-2\ or\ x=1$ .

However, if we plug these solutions back into the original equation, does not create an equality. Therefore, is an extraneous solution.

Compare your answer with the correct one above

Question

Rationalize the denominator:

$\frac{1}{2-\sqrt{3}}$

Answer

The conjugate of $2-\sqrt{3}$ is $2+\sqrt{3}$ .

Now multiply both the numerator and the denominator by $2+\sqrt{3}$

and you get:

$\frac{2+\sqrt{3}}\left ({2-\sqrt{3}} \right )\left ( 2+\sqrt{3} \right )$

$\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right ) =2^{2}-\left ( \sqrt{3} \right )^{2} =4-3 = 1$

Hence we get

$\frac{2+\sqrt{3}}{1} = 2+\sqrt{3}$

Compare your answer with the correct one above

Question

Solve for :

$3^{x-5} = \frac{1}{27}$

Answer

$3^{x-5}=\frac{1}{27}=3^{-3}$

Compare your answer with the correct one above

Question

Solve for .

$\log _{5}x=2$

Answer

$log_5{x} = 2$

$x = 5^{2} = 25$

Compare your answer with the correct one above

Tap the card to reveal the answer