# SOLVING EQUATIONS WITH EXPONENTS WORKSHEET

Problem 1 :

Solve for k :

3x2 ⋅ 2x3 = 6xk

Problem 2 :

If 3x = 10 and 3x-k = 10/27, then solve for k.

Problem 3 :

If 2x+3 - 2x = k(2x), then solve for k.

Problem 4 :

If xac ⋅ xbc = x30, x > 1 and a + b = 5, then solve for c.

Problem 5 :

If x2y3 = 10 and x3y2 = 8, then find the value of x5y5.

Problem 6 :

If x8y7 = 333 and x7y6 = 3, then find the value of xy.

Problem 7 :

If x2 = y3 and x3z = y9, then solve for a.

Problem 8 :

If (√9)-7 ⋅ (√3)-4 = 3k, then solve for k.

Problem 9 :

If x√x = (x√x)x, then solve for x.

Problem 10 :

If 2x = 3y = 6-z, then find the value of

1/x + 1/y + 1/z 3x2 ⋅ 2x3 = 6xk

(3 ⋅ 2)(x⋅ x3) = 6xk

(6)(x2+3) = 6xk

6x5 = 6xk

Divide each side by 6.

x5 = xk

Using laws of exponents, we have

5 = k

So, the value of k is 5.

3x-k = 10/27

Using laws of exponents, we have

3x/3k = 10/27

Substitute 10 for 3x.

10/3k = 10/27

Take reciprocal on both sides.

3k/10 = 27/10

Multiply each side by 10.

3k = 27

3k = 33

Using laws of exponents, we have

k = 3

Using laws of exponents, we have

2x ⋅ 23 - 2x = k(2x)

2x ⋅ 8 - 2x = k(2x)

2x(8 - 1) = k(2x)

2x(7) = k(2x)

Divide each side by 2x.

7 = k

xac ⋅ xbc = x30

Using laws of exponents, we have

xac+bc = x30

ac + bc = 30

Factor.

c(a + b) = 30

Substitute 5 for (a + b).

5c = 30

Divide each side by 5.

c = 6

x2y3 = 10 ----(1)

x3y2 = 8 ----(2)

Multiply (1) and (2) :

(1) ⋅ (2) ----> (x2y3) ⋅ (x3y2) = 10 ⋅ 8

x5y5 = 80

So, the value x5yis 80.

x8y7 = 333 ----(1)

x7y6 = 3 ----(2)

Divide (1) by (2) :

(1) ÷ (2) ----> (x8y7) ÷ (x7y6) = 333 ÷ 3

(x8y7)/(x7y6) = 333/3

Simplify.

x8-7 ⋅ y7-6 = 111

x⋅ y1 = 111

xy = 111

x3z = y9

x3z = y⋅ 3

x3z = (y3)3

Substitute xfor y3.

x3z = (x2)3

x3z = x6

3z = 6

Divide each side by 3.

z = 2

(91/2)-7 ⋅ (31/2)-4 = 3k

(9)-7/2 ⋅ (3)-4/2 = 3k

(32)-7/2 ⋅ 3-2 = 3k

3⋅ (-7/2) ⋅ 3-2 = 3k

3-7 ⋅ 3-2 = 3k

3-7-2 = 3k

3-9 = 3k

k = -9

x√x = (x√x)x

If there is no exponent for any term, we can assume that the exponent of the term is 1.

So, we have

(x√x)1 = (x√x)x

Using the properties of x, we have

1 = x

Let 2x = 3y = 6-z = k.

Then we have

2x = k ----> 2 = k1/x

3y = k ----> 3 = k1/y

6-z = k ----> 6 = k-1/z

Now, we have

⋅ 3 = 6

k1/x ⋅ k1/y  =  k-1/z

Using laws of exponents, we have

k1/x + 1/y = k-1/z

1/x + 1/y = -1/z

1/x + 1/y + 1/z = 0

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