SOLVING LOGARITHMIC EQUATIONS WORKSHEET

Solving Logarithmic Equations Worksheet : 

Worksheet given in this section will be much useful for the students who would like to practice problems on solving logarithmic equations.

Before look at the worksheet, if you would like to know more about logarithms,

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Solving Logarithmic Equations Worksheet - Problems

Problem 1 :

Solve for x :

log₂x  =  1/2

Problem 2 :

Solve for x :

log_1/5x  =  3

Problem 3 :

Solve for x :

log_x125√5  =  7

Problem 4 :

Solve for x :

log_x0.001  =  -3

Problem 5 :

Solve for x : 

log₅(5log₃x)  =  2

Problem 6 :

Solve for x : 

x + 2log₂₇9  =  0

Problem 7 :

If 2logx  =  4log3,  then find the value of x. 

Problem 8 :

If 3x is equal to log(0.3) to the base 9, then find the value of x.  

Problem 9 :

Solve for x :

log₅ √(7x - 4) - 1/2  =  log₅ √(x + 2)

Problem 10 :

Solve for x :

log₃x + log₉x + log₈₁x  =  7/4

Solving Logarithmic Equations Worksheet - Solutions

Problem 1 :

Solve for x :

log₂x  =  1/2

Solution :

log₂x  =  1/2

Convert to exponential form. 

x  =  2^1/2

x  =  √2

Problem 2 :

Solve for x :

log_1/5x  =  3

Solution :

log_1/5x  =  3

Convert to exponential form. 

x  =  (1/5)³

x  =  1³/5³

x  =  1/125

Problem 3 :

Solve for x :

log_x125√5  =  7

Solution :

log_x125√5  =  7

Convert to exponential form. 

125√5  =  x⁷

 5 ⋅ 5 ⋅ 5 ⋅ √5  =  x⁷

Each 5 can be expressed as (√5 ⋅ √5).

Then,

√5 ⋅ √5 ⋅ √5 ⋅ √5 ⋅ √5 ⋅ √5 ⋅ √5  =  x⁷

√5⁷  =  x⁷

Because the exponents are equal, bases can be equated. 

x  =  √5

Problem 4 :

Solve for x :

log_x0.001  =  -3

Solution :

log_x0.001  =  -3

Convert to exponential form. 

0.001  =  x^-3

1/1000  =  1/x³

Take reciprocal on both sides. 

1000  =  x³

10³  =  x³

Because the exponents are equal, bases can be equated. 

10  =  x

Problem 5 :

Solve for x : 

log₅(5log₃x)  =  2

Solution :

log₅(5log₃x)  =  2

Convert to exponential form. 

5log₃x  =  5²

5log₃x  =  25

Divide each side by 5.

log₃x  =  5

Convert to exponential form. 

x  =  3⁵

x  =  243

Problem 6 :

Solve for x : 

x + 2log₂₇9  =  0

Solution :

x + 2log₂₇9  =  0

x  =  -2log₂₇9

x  =  log₂₇9^-2

Convert to exponential form. 

27^x  =  9^-2

(3³)^x  =  (3²)^-2

3^3x  =   3^-4

Because the bases are equal, exponents can be equated. 

3x  =  -4

x  =  -4/3

Problem 7 :

If 2logx  =  4log3,  then find the value of x. 

Solution : 

2logx  =  4log3

Divide each side by 2.

logx  =  (4log3) / 2

logx  =  2log3

logx  =  log3²

logx  =  log9

x  =  9

Problem 8 :

If 3x is equal to log(0.3) to the base 9, then find the value of x.  

Solution : 

From the information given, we have

3x  =  log₉(0.3)

Solve for x.

3x  =  log₉(1/3)

3x  =  log₉1 - log₉3

3x  =  0 - log₉3

3x  =  - log₉3

3x  =  - 1 / log₃9

3x  =  - 1 / log₃3²

3x  =  - 1 / 2log₃3

3x  =  - 1 / 2(1)

3x  =  -1/2

x  =  -1/6

Problem 9 :

Solve for x :

log₅ √(7x - 4) - 1/2  =  log₅ √(x + 2)

Solution :

log₅ √(7x - 4) - 1/2  =  log₅ √(x + 2)

Subtract log₅ √(x + 2) from each side. 

log₅ √(7x - 4) - log₅ √(x + 2) - 1/2  =  0

Add 1/2 to each side. 

log₅ √(7x - 4) - log₅ √(x + 2)  =  1/2

Use quotient rule.

log₅[√(7x - 4) / √(x + 2)]  =  1/2

Convert to exponential form.

[√(7x - 4) / √(x + 2)]  =  5^1/2 

(7x - 4) / (x + 2)  =  √5 

Square each side.  

(7x - 4) / (x + 2)  =  5 

Multiply each side by (x + 2). 

7x - 4  =  5 (x + 2)

7x - 4  =  5x + 10

Subtract 5x from each side.

2x - 4  =  10

Add 4 to each side.

2x  =  14

Divide each side by 2.

x  =  7

Problem 10 :

Solve for x :

log₃x + log₉x + log₈₁x  =  7/4

Solution :

log₃x + log₉x + log₈₁x  =  7/4

(1 / log_x3)  +  (1 / log_x9)  +  (1 / log_x81)  =  7/4

(1 / log_x3)  +  (1 / log_x9)  +  (1 / log_x81)  =  7/4

(1 / log_x3)  +  (1 / log_x3²)  +  (1 / log_x3⁴)  =  7/4

(1 / log_x3)  +  (1 / 2log_x3)  +  (1 / 4log_x3)  =  7/4

(4 / 4log_x3)  +  (2 / 4log_x3)  +  (1 / 4log_x3)  =  7/4

(4 + 2 + 1) / 4log_x3  =  7/4

7 / 4log_x3  =  7/4

Multiply each side by 4/7.

1 / log_x3  =  1

log₃x  =  1

Convert to exponential form. 

x  =  3¹

x  =  3

After having gone through the stuff given above, we hope that the students would have understood how to solve logarithmic equations. 

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