**Middle School Number Sense Lesson 38: Solving for X (with Exponents)**

Earlier this week we learned how to solve for X (without exponents)--in my

**SolveX**post. You may want to review that lesson if you haven't already. Today we will add a little complexity and introduce exponents. This concept appeared

**9 times**this year, with a median placement at

**question # 39**.

**Number Dojo Level: 243**

There is quite a variety of question types within this topic. Since I have already demonstrated the basic steps in Wednesday's post, I will jump straight to some examples.

**Example 1:**

**If 5x + 11 = 21, then x**

^{3}= ___1.

**11**is the constant. Subtract 11 from both sides of the equation. 5x + 11

**- 11**= 21

**- 11**. You are left with

**5x = 10**.

2.

**5**is the coefficient. Divide both sides of the equation by 5. 5x

**÷ 5**= 10

**÷ 5**. You are left with

**x = 2**.

3. The question asked for

**x**. x

^{3}^{3}= 2

^{3}=

**8**.

Example 2:

Example 2:

**If x**

^{3}= 125, then 7x + 1 = ___1. Take the cube root of both sides to get

**x = 5**.

2. Now substitute

**5**back into the equation for x. 7x + 1 = 7(5) + 1 =

**36**.

**Example 3:**

**If x**

^{2}= 121 and x < 0, then x = ___1. Take the square root of both sides to get

**x = 11**.

2. Since x is less than 0, the answer has to be

**-11**.

**Example 4:**

**If 4**

^{2}+ 8^{2}+ 16^{2}= (4)^{2}k, then k = ___1. Factor 4

^{2}out of each term.

**(4**

^{2})(1 + 2^{2}+ 4^{2}) = (4)^{2}k.2. Simplify within the parentheses. (1 + 2

^{2}+ 4

^{2}= 1 + 4 + 16 =

**21**.

3. Plug this back in to get (4

^{2})(21) = (4)

^{2}k.

4. Divide both sides by 4

^{2}. You are left with

**21 = k**.

**Example 5:**

**If x - y = 12 and x + y = 18, then x**

^{2}- y^{2}= ___1. This is an extension of the

**DiffSquares1**and

**DiffSquares2**concepts, which teach us that

**x**.

^{2}- y^{2}= (x + y)(x - y)2. Simply multiply 12 by 18 to get

**216**. (We can do that easily because we've mastered the

**Mult10s,U+10**concept).

**Example 6:**

**If 2**

^{3}x 4^{4}= 2^{k}, then k = ___1. Use the rules of exponents to change the left side to:

**2**.

^{3}x (2^{2})^{4}2. Simplify. 2

^{3}x (2

^{2})

^{4}= 2

^{3}x 2

^{(2 x 4)}= 2

^{3}x 2

^{8}=

**2**.

^{11}3. Plug this back in. 2

^{11}= 2

^{k}, so

**k = 11**.

**Here's a free worksheet to help you practice SolveXExpo:**

solvexexpo.pdf |

**Up Next for Middle School: EstSquareRoot**