SOLVING EXPONENTIAL EQUATIONS WITH DIFFERENT BASES WORKSHEET

Solve for x in each of the following exponential equations.

Problem 1 :

2^x = 4

Problem 2 :

4^x = 1/2

Problem 3 :

9^x = 1/27

Problem 4 :

36^x = 1/6

Problem 5 :

5^x = 1

Problem 6 :

8^{x + 1} = 16

Problem 7 :

8^{2 - x} = 1/64

Problem 8 :

(1/2)^{x - 1} = 4

Problem 9 :

(1/3)^{x - 2} = 1/9

Problem 10 :

2^x - 1 = 5

Problem 11 :

5^{x - 1} - 2^x = 0

Problem 12 :

5^x = (√25)^-7 ⋅ (√5)^-5)

Problem 13 :

4^x = 7(2^x) + 18

Answers

1. Answer :

2^x = 4

2^x = 2²

x = 2

2. Answer :

4^x = 1/2

(2²)^x = 1/2¹

2^2x = 2^-1

2x = -1

Divide both sides by 2.

x = -1/2

3. Answer :

9^x = 1/27

(3²)^x = 1/3³

3^2x = 3^-3

2x = -3

Divide both sides by 2.

x = -3/2

4. Answer :

36^x = 1/6

(6²)^x = 1/6¹

6^2x = 6^-1

2x = -1

Divide both sides by 2.

x = -1/2

5. Answer :

5^x = 1

5^x = 5⁰

x = 0

6. Answer :

8^{x + 1} = 16

(2³)^{x + 1}  =  2⁴

2^{3x + 1}  =  2⁴

3x + 1 = 4

Subtract 1 from both sides.

3x = 3

Divide both sides by 3.

x = 1

7. Answer :

8^{2 - x} = 1/64

8^{2 - x} = 1/8²

8^{2 - x} = 8^-2

2 - x = -2

Subtract 2 from both sides.

-x = -4

Multiply both sides by -1.

x = 4

8. Answer :

(1/2)^{x - 1} = 4

(2^-1)^{x - 1} = 2²

2^{-1(x - 1)} = 2²

2^{-x + 1} = 2²

-x + 1 = 2

Subtract 1 from both sides.

-x = 1

Multiply both sides by -1.

x = -1

9. Answer :

(1/3)^{x - 2} = 1/9

(3^-1)^{x - 2} = 3²

3^{-1(x - 2)} = 3²

3^{-x + 2} = 3²

-x + 2 = 9

Subtract 2 from both sides.

-x = 7

Multiply both sides by -1.

x = -7

10. Answer :

2^x - 1 = 5

Add 1 to both sides.

2^x = 5

In the given equation, we have the base 2 on the left side and 5 on the right side is not a power of 2. Since we don't have the same base on both sides, we can take logarithm with any base on both sides and solve for x.

log(2^x) = log6

Using power rule of logarithm,

xlog2 = log6

Divide both sides by log2.

11. Answer :

5^{x - 1} - 2^x = 0

Add 2^x to both sides.

5^{x - 1} = 2^x

In the equation above, 5 is not a power of 2 and also 2 is not a power of 5. Since we don't have the same base on both sides, we can take logarithm with any base on both sides and solve for x.

log(5^{x - 1}) = log(2^x)

Using power rule of logarithm,

(x - 1)log5 = xlog2

Using distributive property,

xlog5 - log5 = xlog2

Subtract xlog2 from both sides.

xlog5 - log5 - xlog2 = 0

Add log5 to both sides.

xlog5 - xlog2 = log5

Factor.

x(log5 - log)2 = log5

Divide both sides by (log5 - log2).

12. Answer :

5^x = (√25)^-7 ⋅ (√5)^-5

5^x = 5^-7 ⋅ (5^1/2)^-5

5^x = 5^-7 ⋅ 5^-5/2

5^x = 5^{-7 - 5/2}

5^x = 5^-19/2

Equate the exponents.

x = -19/2

13. Answer :

4^x = 7(2^x) + 18

(2²)^x = 7(2^x) + 18

(2^x)² = 7(2^x) + 18

(2^x)² - 7(2^x) - 18 = 0

Let a = 2^x.

a² - 7a - 18 = 0

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

a - 9 = 0  or  a + 2 = 0

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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