How to multiply exponents - PSAT Math

The three two multiply to become 8 and the powers of ten can be added to become 1021.

c4 is equal to (c2)(c2).

We know c2 = 15. Plugging in this value gives us c4 = (15)(15) = 225.

and can be multiplied together to give you which is the first part of our answer. When you multiply exponents with the same base (in this case, ), you add the exponents. In this case, should give us because . The answer is

(x3y6z)(x2yz3)

The paraentheses are irrelevant. Rearrange to combine like terms.

x3x2y6y1z1z3

When you multiply variables with exponents, simply add the exponents together.

x3+2 y6+1 z1+3

x5y7z4

The question tells us that 22_a_ ( 22_b_ )= 16.

We can rewrite 16 as 24, giving us 22_a_ ( 22_b_ )= 24.

When terms with the same base are multipled, their exponents can be added:

2(2_a_ +2_b_) = 24

Since the base is the same on both sides of the equation, we can equate the exponents:

2_a_ +2_b_ = 4

2(a + b) = 4

a + b = 2

1. Solve for x in 3x = 27. x = 3 because 3 * 3 * 3 = 27.
2. Since x = 3, one can substitute x for 3 in 22x
3. Now, the expression is 22*3
4. This expression can be interpreted as 22 * 22 * 22. Since 22 = 4, the expression can be simplified to become 4 * 4 * 4 = 64.
5. You can also multiply the powers to simplify the expression. When you multiply the powers, you get 26, or 2 * 2 * 2 * 2 * 2 * 2
6. 26 = 64.

In order to solve this equation, we first need to find a common base for the exponents. We know that 23 = 8 and 24 = 16, so it makes sense to use 2 as a common base, and then rewrite each side of the equation as a power of 2.

8x-3 = (23)x-3

We need to remember our property of exponents which says that (ab)c = abc.

Thus (23)x-3 = 23(x-3) = 23x - 9.

We can do the same thing with 164-x.

164-x = (24)4-x = 24(4-x) = 216-4x.

So now our equation becomes

23x - 9 = 216-4x

In order to solve this equation, the exponents have to be equal.

3x - 9 = 16 - 4x

Add 4x to both sides.

7x - 9 = 16

Add 9 to both sides.

7x = 25

Divide by 7.

x = 25/7.

We can start by rewriting 411 as 4 * 410. This will allow us to collect the like terms 410 into a single term.

410 + 410 + 410 + 410 + 411

= 410 + 410 + 410 + 410 + 4 * 410

= 8 * 410

Because the answer choices are written with a base of 2, we need to rewrite 8 and 4 using bases of two. Remember that 8 = 23, and 4 = 22.

8 * 410

= (23)(22)10

We also need to use the property of exponents that (ab)c = abc. We can rewrite (22)10 as 22x10 = 220.

(23)(22)10

= (23)(220)

Finally, we must use the property of exponents that ab * ac = ab+c.

(23)(220) = 223

The answer is 223.

3 + 3_n_+3 = 81

In this equation, there is a common factor of 3, which can be factored out.

Thus, 3(1 + 3_n_+2) = 81

Note: when 3 is factored out of 3_n_+3, the result is 3_n_+2 because (3_n_+3 = 31 * 3_n_+2). Remember that exponents are added when common bases are multiplied. Also remember that 3 = 31.

3(1 + 3_n_+2) = 81

(1 + 3_n_+2) = 27

3_n_+2 = 26

Note: do not solve for n individually. But rather seek to solve what the problem asks for, namely 3n+2.

First, let us use the definiton of f(x) to find f(10).

f(x) = (2 – x)(x/3)

f(10) = (2 – 10)(10/3)

= (–8)(10/3)

In order to evaluate the above expression, we can make use of the property of exponents that states that abc = (ab)c = (ac)b.

(–8)(10/3) = (–8)10(1/3) = ((–8)(1/3))10.

(–8)(1/3) requires us to take the cube root of –8. The cube root of –8 is –2, because (–2)3 = –8.

Let's go back to simplifying ((–8)(1/3))10.

((–8)(1/3))10 = (–2)10 = f(10)

We are asked to find n such that 4n = (–2)10. Let's rewrite 4n with a base of –2, because (–2)2 = 4.

4n = ((–2)2)n = (–2)2n = (–2)10

In order to (–2)2n = (–2)10, 2n must equal 10.

2n = 10

Divide both sides by 2.

n = 5.

The answer is 5.

For this problem, we need to make use of the property of exponents, which states that (xy)z = xyz.

We are given a2 but are asked to find a6.

Let's raise both sides of the equation to the third power, so that we will end up with a6 on the left side.

(a2)3 = (b3)3

Now, according to the property of exponents mentioned before, we can multiply the exponents.

a(2*3) = b(3*3)

a6 = b9

The answer is b9.

In order to solve this equation, we are going to need to use a common base. Because 2, 4, 8, and 16 are all powers of 2, we can rewrite both sides of the equation using 2 as a base. Since 22 = 4, 23 = 8, and 24 = 16, we can rewrite the original equation as follows:

2_n *_ 4_n *_ 8_n *_ 16 = 2_–n *_ 4_–n *_ 8_–n_

2_n_(22)n(23)n(24) = 2_–n_(22)– n(23)– n

Now, we will make use of the property of exponents which states that (ab)c = abc.

2_n_(22_n_)(23_n_)(24) = 2_–n_(2_–2_n)(2_–3_n)

Everything is now written as a power of 2. We can next apply the property of exponents which states that abac = a b+c.

2(n+2_n_+3_n_+4) = 2(–n + –2_n_ + –3_n_)

We can now set the exponents equal and solve for n.

n + 2_n_ + 3_n_ + 4 = –n + –2_n_ + –3_n_

Let's combine the n's on both sides.

6_n_ + 4 = –6_n_

Add 6_n_ to both sides.

12_n_ + 4 = 0

Subtract 4 from both sides.

12_n_ = –4

Divide both sides by 12.

n = –4/12 = –1/3

The answer is –1/3.

First, we need to solve 1252_x_–4 = 6257–x . When solving equations with exponents, we usually want to get a common base. Notice that 125 and 625 both end in five. This means they are divisible by 5, and they could be both be powers of 5. Let's check by writing the first few powers of 5.

51 = 5

52 = 25

53 = 125

54 = 625

We can now see that 125 and 625 are both powers of 5, so let's replace 125 with 53 and 625 with 54.

(53)2_x_–4 = (54)7–x

Next, we need to apply the rule of exponents which states that (ab)c = abc .

53(2_x_–4) = 54(7–x)

We now have a common base expressed with one exponent on each side. We must set the exponents equal to one another to solve for x.

3(2_x –_ 4) = 4(7 – x)

Distribute the 3 on the left and the 4 on the right.

6_x_ – 12 = 28 – 4_x_

Add 4_x_ to both sides.

10_x_ – 12 = 28

Add 12 to both sides.

10_x_ = 40

Divide by 10 on both sides.

x = 4

However, the question asks us for the largest prime factor of x. The only factors of 4 are 1, 2, and 4. And of these, the only prime factor is 2.

The answer is 2.

When an exponent is raised to a power, we multiply. But when two exponents with the same base are multiplied, we add them. So (_x_3)2 = _x_3*2 = _x_6. Then (_x_3)2 * _x–_2 = _x_6 * _x–_2 = _x_6 – 2 = _x_4.

We can break up the equation into two smaller equations involving only x and y. Then, once we solve for x and y, we can find the value of $x^3y$^4.

must be negative because it has an odd power and and have even powers above. But and could be positive or negative, so none of the scenarios has to be true.

To find the answer we can apply the equation of population where is the number of hours.

$(6x^{2}$$)^{2}$ = $6^{2}$$x^{4}$ = $36x^{4}$

$36x^{4}$times $y^{2}$ = $36x^{4}$$y^{2}$

We cannot combine $36x^{4}$$y^{2}$ with xyz, so $(6x^{2}$$)^{2}$times $y^{2}$ + xyz $=36x^{4}$$y^{2}$+xyz.

Use the distributive property: a(b+c)=ab+ac. When we multiply variables with exponents, we keep the same base and add the exponents: $a^{m}$$a^{n}$ = $a^{m+n}$