SOLVING EXPONENTIAL EQUATIONS WITH LOGARITHMS WORKSHEET

Problem 1 :

Find the value of x in the following equation using conversion between exponential and logarithm.

3^x = 27

Problem 2 :

Find the value of x in the following equation by taking logarithm on both sides.

3^x = 27

Problem 3 :

Find the value of x in the following equation using conversion between exponential and logarithm.

2^{x - 5} = 1/16

Problem 4 :

Find the value of x in the following equation by taking logarithm on both sides.

2^{x - 5} = 1/16

Problem 5 :

Solve for x :

2^x - 1 = 5

Problem 6 :

Solve for x :

5^{x - 1} - 2^x = 0

1. Answer :

3^x = 27

Convert the above equation to logarithm.

x = log₃27

x = log₃(3³)

Using the power rule of logarithm,

x = 3log₃3

x = 3(1)

x = 3

2. Answer :

3^x = 27

In the given equation, we have the base 3 on the left side and 27 on the right side can be written as 3³. Since we have the same base 3 on both sides, we can take logarithm with the base 3 on both sides and solve for x.

log₃3^x = log₃27

log₃(3^x) = log₃(3³)

xlog₃3 = 3log₃3

x(1) = 3(1)

x = 3

3. Answer :

2^{x - 5} = 1/16

Convert the above equation to logarithm.

x - 5 = log₂(1/16)

x - 5 = log₂(1/2⁴)

x - 5 = log₂(2^-4)

Using the power rule of logarithm,

x - 5 = -4log₂2

x - 5 = -4(1)

x - 5 = -4

Add 5 to both sides.

x = 1

4. Answer :

2^{x - 5} = 1/16

In the given equation, we have the base 2 on the left side and 1/16 on the right side can be written as 2^-4. Since we have the same base 2 on both sides, we can take logarithm with the base 3 on both sides and solve for x.

log₂(2^{x - 5}) = log₂(1/16)

log₂(2^{x - 5}) = log₂(1/2⁴)

log₂(2^{x - 5}) = log₂(2^-4)

Using the power rule of logarithm,

(x - 5)log₂2 = -4log₂2

(x - 5)(1) = -4(1)

x - 5 = -4

Add 5 to both sides.

x = 1

5. 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.

6. 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).

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