Exponents - PSAT Math

Card 0 of 1029

Question

Simplify_exponent_7-11-13

Answer

Simplify_exponent_2_7-11-13

Simplify_exponent_4_7-11-13

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Question

54 / 25 =

Answer

25 = 5 * 5 = 52. Then 54 / 25 = 54 / 52.

Now we can subtract the exponents because the operation is division. 54 / 52 = 54 – 2 = 52 = 25. The answer is therefore 25.

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Question

Com_exp_1

Which of the following lists the above quantities from least to greatest?

Answer

Com_exp_2

Com_exp_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

  1. 5. Simplify the problem (x4y2/x5)3

Answer

Properties of exponents suggests that when multiplying the same base, add the exponents, when dividing, subtract the exponents on bottom from those on top, and when raising an exponent to another power, multiply the exponents. Remember that (x4/x5) = x–1 = 1/x; Still using order of operations (PEMDAS) we get the following:(x4y2/x5)3= (y2/x)3 = y6/(x3).

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Question

If , 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 , what is the value of ?

Answer

Since we have two ’s in we will need to combine the two terms.

For this can be rewritten as

So we have .

Or

Divide this by :

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.

We can rewrite the original equation in these terms.

Simplify exponents.

Finally, combine terms.

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 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:

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

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

Answer

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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.

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

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Question

Simplify:

Answer

Similarly

So

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