SOLVING EXPONENTIAL EQUATIONS USING EXPONENT PROPERTIES WORKSHEET

Solve for x in each of the following :

Problem 1 :

4^x = 64

Problem 2 :

9^{x - 1} = 27

Problem 3 :

8^x = 1

Problem 4 :

Problem 5 :

9 ⋅ 2^x = 72

Problem 6 :

Problem 7 :

Problem 8 :

Problem 9 :

3^{x - 1}⋅ 9^{2x + 1} = 243

Problem 10 :

9^x = 7(3^x) + 18

Answers

1. Answer :

4^x = 64

4^x = 4³

x = 3

2. Answer :

9^{x - 1} = 27

(3²)^{x - 1} = 3³

3^{2(x - 1)} = 3³

3^{2x - 2} = 3³

2x - 2 = 3

2x = 5

3. Answer :

8^x = 1

8^x = 8⁰

x = 0

4. Answer :

5^{x - 2} = 5^-2

x - 2 = -2

x = 0

5. Answer :

9 ⋅ 2^x = 72

Divide both sides by 9.

2^x = 8

2^x = 2³

x = 3

6. Answer :

(3³)^{x - 1} = (3^-1)^{1 - 2x}

3^{3(x - 1)} = 3^{-1(1 - 2x)}

3^{3x - 3} = 3^{-1 + 2x}

3x - 3 = -1 + 2x

x - 3 = -1

x = 2

7. Answer :

5^{2(x + 2)} = 5^-3

5^{2x + 4} = 5^-3

2x + 4 = -3

2x = -7

8. Answer :

2^{x + 1} ⋅ (2²)^x = (2^-1)^{x + 1}

2^{x + 1} ⋅ 2^2x = 2^{-(x + 1)}

2^{x + 1 + 2x} = 2^{-x - 1}

2^{3x + 1} = 2^{-x - 1}

3x + 1 = -x - 1

4x = -2

9. Answer :

3^{x - 1}⋅ 9^{2x + 1} = 243

3^{x - 1}⋅ (3²)^{2x + 1} = 3⁵

3^{x - 1}⋅ 3^{2(2x + 1)} = 3⁵

3^{x - 1}⋅ 3^{4x + 2} = 3⁵

3^{x - 1 + 4x + 2} = 3⁵

3^{5x + 1} = 3⁵

5x + 1 = 5

5x = 4

10. Answer :

9^x = 7(3^x) + 18

(3²)^x = 7(3^x) + 18

(3^x)² = 7(3^x) + 18

Let y = 3^x.

y² = 7y + 18

y² - 7y - 18 =0

y² - 9y + 2y - 18 =0

y(y - 9) + 2(y - 9) =0

(y - 9)(y + 2) = 0

y - 9 = 0  or  y + 2 = 0

y = 9  or  y = -2

In 3^x, whatever real value (positive or negative or zero) we substitute for x, 3^x can never be negative. So we can ignore the equation 3^x = -2.

Therefore,

x = 2

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