SOLVING EQUATIONS WITH EXPONENTS WORKSHEET

Problem 1 :

Solve for k :

3x² ⋅ 2x³ = 6x^k

Problem 2 :

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

Problem 3 :

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

Problem 4 :

If x^ac ⋅ x^bc = 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 x^3z = y⁹, then solve for a.

Problem 8 :

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

Problem 9 :

If x√x = (x√x)^x, then solve for x.

Problem 10 :

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

¹⁄ₓ + ¹⁄y + ¹⁄z

Answers

1. Answer :

3x² ⋅ 2x³ = 6x^k

(3 ⋅ 2)(x² ⋅ x³) = 6x^k

(6)(x²⁺³) = 6x^k

6x⁵ = 6x^k

Divide each side by 6.

x⁵ = x^k

Using laws of exponents, we have

5 = k

So, the value of k is 5.

2. Answer :

Substitute 10 for 3x.

Multiply each side by 10.

3^k = 27

3^k = 3³

Using laws of exponents, we have

k = 3

3. Answer :

Using laws of exponents, we have

2^x ⋅ 2³ - 2^x = k(2^x)

2^x ⋅ 8 - 2^x = k(2^x)

2^x(8 - 1) = k(2^x)

2^x(7) = k(2^x)

Divide each side by 2^x.

7 = k

4. Answer :

x^ac ⋅ x^bc = x³⁰

Using laws of exponents, we have

x^ac+bc = x³⁰

ac + bc = 30

Factor.

c(a + b) = 30

Substitute 5 for (a + b).

5c = 30

Divide each side by 5.

c = 6

5. Answer :

x²y³ = 10 ----(1)

x³y² = 8 ----(2)

Multiply (1) and (2) :

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

x⁵y⁵ = 80

So, the value x⁵y⁵ is 80.

6. Answer :

x⁸y⁷ = 333 ----(1)

x⁷y⁶ = 3 ----(2)

Divide (1) by (2) :

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

Simplify.

x^8-7 ⋅ y^7-6 = 111

x¹ ⋅ y¹ = 111

xy = 111

7. Answer :

x^3z = y⁹

x^3z = y^{3 ⋅ 3}

x^3z = (y³)³

Substitute x² for y³.

x^3z = (x²)³

x^3z = x⁶

3z = 6

Divide each side by 3.

z = 2

8. Answer :

(9^1/2)^-7 ⋅ (3^1/2)^-4 = 3^k

(9)^-7/2 ⋅ (3)^-4/2 = 3^k

(3²)^-7/2 ⋅ 3^-2 = 3^k

3^{2 ⋅ (-7/2)} ⋅ 3^-2 = 3^k

3^-7 ⋅ 3^-2 = 3^k

3^-7-2 = 3^k

3^-9 = 3^k

k = -9

9. Answer :

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

10. Answer :

Let 2^x = 3^y = 6^-z = k.

Then we have

2^x = k ----> 2 = k^1/x

3^y = k ----> 3 = k^1/y

6^-z = k ----> 6 = k^-1/z

Now, we have

2 ⋅ 3 = 6

k^1/x ⋅ k^1/y  =  k^-1/z

Using laws of exponents, we have

k^{1/x + 1/y} = k^-1/z

¹⁄ₓ + ¹⁄y = -¹⁄z

Add 1/z to each side.

¹⁄ₓ + ¹⁄y + ¹⁄z = 0

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