Exponents - PSAT Math

Card 0 of 20

Question

If $3^{4x+6}=27^{2x}$ , what is the value of ?

Answer

Using exponents, 27 is equal to 33. So, the equation can be rewritten:

34_x_ + 6 = (33)2_x_

34_x_ + 6 = 36_x_

When both side of an equation have the same base, the exponents must be equal. Thus:

4_x_ + 6 = 6_x_

6 = 2_x_

x = 3

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Question

If _a_2 = 35 and _b_2 = 52 then _a_4 + _b_6 = ?

Answer

_a_4 = _a_2 * _a_2 and _b_6= _b_2 * _b_2 * _b_2

Therefore _a_4 + _b_6 = 35 * 35 + 52 * 52 * 52 = 1,225 + 140,608 = 141,833

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Question

If $9^{(x + 5)}+3^{2(x+5)}=162$ , what is the value of ?

Answer

Since we have two ’s in $9^{(x + 5)} + 3^{2(x + 5)}$ we will need to combine the two terms.

For $3^{2(x + 5) }$ this can be rewritten as

$(3^2)^{ (x + 5)} = 9^ {(x + 5)}$

So we have $9^{ (x + 5) }+ 9^{ (x + 5)} = 162$ .

Or $2 (9^{ (x + 5)}) = 162$

Divide this by : $9^{ (x + 5)} = 81 = 9^ 2$

Thus or

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

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Question

Solve for x.

23 + 2x+1 = 72

Answer

The answer is 5.

8 + 2x+1 = 72

2x+1 = 64

2x+1 = 26

x + 1 = 6

x = 5

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Question

What is the value of such that ?

Answer

We can solve by converting all terms to a base of two. 4, 16, and 32 can all be expressed in terms of 2 to a standard exponent value.

$4=2^2\rightarrow 4^2=(2^2)^2$

$16=2^4\rightarrow 16^3=(2^4)^3$

$32=2^5\rightarrow 32^2=(2^5)^2$

We can rewrite the original equation in these terms.

Simplify exponents.

$2^n=(2^4)(2^{12})(2^{10})$

Finally, combine terms.

$2^n=2^{4+12+10}=2^{26}$

From this equation, we can see that .

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Question

Which of the following is eqivalent to 5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) , where b is a constant?

Answer

We want to simplify 5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) .

Notice that we can collect the –5(b–1) terms, because they are like terms. There are 5 of them, so that means we can write –5(b–1) – 5(b–1) – 5(b–1) – 5(b–1) – 5(b–1) as (–5(b–1))5.

To summarize thus far:

5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–1) = 5_b +(–5(_b–_1))5

It's important to interpret –5(b–1) as (–1)5(b–1) because the –1 is not raised to the (b – 1) power along with the five. This means we can rewrite the expression as follows:

5_b_ +(–5(b–1))5 = 5_b_ + (–1)(5(b–1))(5) = 5_b_ – (5(b–1))(5)

Notice that 5(b–1) and 5 both have a base of 5. This means we can apply the property of exponents which states that, in general, abac = a b+c. We can rewrite 5 as 51 and then apply this rule.

5_b_ – (5(_b–1))(5) = 5_b – (5(_b–1))(51) = 5_b – 5(_b–_1+1)

Now, we will simplify the exponent b – 1 + 1 and write it as simply b.

5_b_ – 5(b–1+1) = 5_b* – 5_b* = 0

The answer is 0.

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Question

If $\dpi{100} \small r$ and $\dpi{100} \small s$ are positive integers, and $\dpi{100} \small 25\left ( 5^{r} \right )=5^{s-2}$ , then what is $\dpi{100} \small s$ in terms of $\dpi{100} \small r$ ?

Answer

$\dpi{100} \small 25\left ( 5^{r} \right )$ is equal to $5^{2}\left ( 5^{r}\right )$ which is equal to $\dpi{100} \small \left ( 5^{r+2} \right )$ . If we compare this to the original equation we get $\dpi{100} \small r+2=s-2\rightarrow s=r+4$

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Question

Solve for $x$ :

$4^{4}+4^{4}+4^{4}+4^{4}=2^{x}$

Answer

Combining the powers, we get $1024=2^{x}$ .

From here we can use logarithms, or simply guess and check to get $x=10$ .

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Question

If $x^2=11$ , then what does $x^4$ equal?

Answer

$x^4=(x^2)(x^2)=11\cdot 11=121$

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Question

Simplify. All exponents must be positive.

$\left ( x^{-2}y^{3} \right )\left ( x^{5}y^{-4} \right )$

Answer

Step 1: $\left ( x^{-2}x^{5} \right )= x^{3}$

Step 2: $\left ( y^{3}y^{-4} \right )= y^{-1}= \frac{1}{y}$

Step 3: (Correct Answer): $\frac{x^{3}}{y}$

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Question

Simplify. All exponents must be positive.

$\frac{x^{-3}y^{5}}{x^{2}y^{-1}}$

Answer

Step 1: $\frac{y^{5}}{\left ( x^{3}x^{2} \right )\left \right )y^{-1}}$

Step 2: $\frac{\left ( y^{5}y^{1} \right )}{x^{3}x^{2}}$

Step 3: $\frac{y^{6}}{x^{5}}$

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Question

$\frac{\left ( -11 \right )^{-8}}{\left ( -11\right )^{12}}$

Answer must be with positive exponents only.

Answer

Step 1: $\frac{1}{\left ( -11 \right )^{12}\left ( -11 \right )^{8}}$

Step 2: The above is equal to $\frac{1}{\left ( -11 \right )^{20}}$

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Question

Evaluate:

$-\left ( -3 \right )^{0}-\left ( -3^{0} \right )$

Answer

$-\left ( -3 \right )^{0}= -1$

$(-3^{0})=-1$

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Question

Simplify:

$3\sqrt{18} + 2\sqrt{50}$

Answer

$3\sqrt{18} = 3\sqrt{2\star 3^{2}} = 9\sqrt{2}$

Similarly $2\sqrt{50} = 2\sqrt{2\star 5^{2}} = 10\sqrt{2}$

So $9\sqrt{2} + 10\sqrt{2}= 19\sqrt{2}$

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Question

How many of the following base ten numbers have a base five representation of exactly four digits?

(A)

(B)

(C)

(D)

Answer

A number in base five has powers of five as its place values; starting at the right, they are $1, 5 ^{1} = 5, 5 ^{2} = 25, 5 ^{3} = 125,5 ^{4} = 125$

The lowest base five number with four digits would be

$1000 _{\textrm{five}} = 1 \times 5^{3} = 125$ in base ten.

The lowest base five number with five digits would be

$10000 _{\textrm{five}} = 1 \times 5^{4} = 625$ in base ten.

Therefore, a number that is expressed as a four-digit number in base five must fall in the range

$125 \leq N \leq 624$

Three of the four numbers - all except 100 - fall in this range.

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Question

Simplify by rationalizing the denominator:

$\frac{20 + 6i}{8i}$

Answer

Multiply both numerator and denominator by :

$\frac{20 + 6i}{8i}$

$=\frac{\left (20 + 6i \right ) \cdot i}{8i \cdot i}$

$=\frac{ 20i + 6i^{2} }{8i ^{2}}$

$=\frac{ 20i + 6(-1) }{8 (-1)}$

$=\frac{ 20i -6 }{-8}$

$=\frac{ -6+ 20i }{-8}$

$=\frac{ -6 }{-8} + \frac{ 20i }{-8}$

$=\frac{ 3 }{4} - \frac{ 5}{2}i$

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Question

Divide:

$-100 \div 5i$

Answer

Rewrite in fraction form, and multiply both numerator and denominator by :

$-100 \div 5i$

$= \frac{-100}{5i}$

$= \frac{-100 \cdot i }{5i \cdot i }$

$= \frac{-100 i }{5i ^{2}}$

$= \frac{-100 i }{5 (-1)}$

$= \frac{-100 i }{-5}$

$= \frac{-100 }{-5}\cdot i$

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Question

Simplify by rationalizing the denominator:

$\frac{ 25- 12i}{ - 10i}$

Answer

Multiply both numerator and denominator by :

$\frac{ 25- 12i}{ - 10i}$

$=\frac{\left ( 25- 12i \right )\cdot i}{ - 10i \cdot i }$

$=\frac{ 25i- 12i ^{2} }{ - 10i ^{2} }$

$=\frac{ 25i- 12 (-1)}{ - 10 (-1) }$

$=\frac{ 25i+ 12}{ 10}$

$=\frac{12+ 25i }{ 10}$

$=\frac{12 }{ 10} + \frac{ 25i }{ 10}$

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

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Question

Simplify by rationalizing the denominator:

$\frac{ 21i}{3+ i\sqrt{5}}$

Answer

Multiply the numerator and the denominator by the complex conjugate of the denominator, which is $3- i\sqrt{5}$ :

$\frac{ 21i}{3+ i\sqrt{5}}$

$= \frac{ 21i\left (3- i\sqrt{5} \right )}{\left (3+ i\sqrt{5} \right )\left (3- i\sqrt{5} \right )}$

$= \frac{21i \cdot 3- 21i \cdot i\sqrt{5} }{ 3^{2}+ \left ( \sqrt{5} \right )^{2}}$

$= \frac{ 63i - 21i^{2} \sqrt{5}}{ 9+5}$

$= \frac{ 63i - 21 (-1)\sqrt{5}}{ 14}$

$= \frac{ 21 \sqrt{5}+ 63i }{ 14}$

$= \frac{ 21 \sqrt{5} }{ 14}+ \frac{63 i}{ 14}$

$= \frac{ 3\sqrt{5} }{ 2} + \frac{ 9 }{ 2} i$

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Question

Simplify by rationalizing the denominator:

$\frac{2+3i}{4+ 3i}$

Answer

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

$\frac{2+3i}{4+ 3i}$

$=\frac{\left (2+3i \right )\left (4-3i \right )}{\left (4+ 3i \right )\left (4-3i \right )}$

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

$=\frac{8 - 6i +12i - 9 i ^{2}}{16+ 9}$

$=\frac{8 - 6i +12i - 9(-1)}{25}$

$=\frac{8 - 6i +12i +9}{25}$

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

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

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