SOLVING LOGARITHMIC EQUATIONS

Problem 1 :

Solve for x :

log2x = 1/2

Solution :

log2x = 1/2

Convert to exponential form. 

x = 21/2

x = √2

Problem 2 :

Solve for x :

log1/5x = 3

Solution :

log1/5x = 3

Convert to exponential form. 

x = (1/5)3

x = 13/53

x = 1/125

Problem 3 :

Solve for y :

log3y = -2

Solution :

log3y = -2

Convert to exponential form. 

y = 3-2

y = 1/32

y = 1/9

Problem 4 :

Solve for x :

logx125√5 = 7

Solution :

logx125√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

√5= x7

Because the exponents are equal, bases can be equated. 

x = √5

Problem 5 :

Solve for x :

logx0.001 = -3

Solution :

logx0.001 = -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 6 :

Solve for x : 

x + 2log279 = 0

Solution :

x + 2log279 = 0

x = -2log279

x = log279-2

Convert to exponential form. 

27x = 9-2

(33)= (32)-2

33x = 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 the following equation :

log4(x + 4) + log48 = 2

Solution :

log4(x + 4) + log48 = 2

Combine the two terms on the left side.

log4[8 ⋅ (x + 4)] = 2

log4(8x + 32) = 2

8x + 32 = 42

8x + 32 = 16

Subtract by 32 from both sides

8x = -16

Divide both sides by 8.

x = -2

Problem 10 :

Solve the following equation :

log6(x + 4) - log6(x - 1) = 1

Solution :

log6(x + 4) - log6(x - 1) = 1

Combine the two terms on the left side

log6[(x + 4)/(x - 1)] = 1

(x + 4)/(x - 1) = 61

(x + 4)/(x - 1) = 6

x + 4 = 6(x - 1)

x + 4 = 6x - 6

Subtract 6x from both sides.

x - 6x + 4 = -6

-5x + 4 = -6

Subtract 4 from both sides.

-5x = -6 - 4

-5x = -10

Divide both sides by -5.

x = 2

Problem 11 :

Solve the following equation :

log2x + log4x + log8x = 11/6

Solution :

log2x + log4x + log8x = 11/6

(1/logx2) + (1/logx4) + (1/logx8) = 11/6

(1/logx2) + (1/logx22) + (1/logx23) = 11/6

(1/logx2) + (1/2logx2) + (1/3logx2) = 11/6

(1/logx2) (1 + 1/2 + 1/3) = 11/6

(1/logx2)(11/6) = 11/6

1/logx2 = 1

1 = logx

x = 2

Problem 12 :

Given that

logx = m + n

logy = m – n

Find the value of log(10x/y2) in terms of m and n.

Solution :

log(10x/y2) = log10x - 1ogy2

= log10 + logx - 2logy

= 1 + logx - 2logy

Substitute.

= 1 + (m + n) - 2(m - n)

= 1 + m + n - 2m + 2n

= 1 - m + 3n

Problem 13 :

Given that

logx + logy = log(x + y)

Solve for y in terms of x.

Solution :

logx + logy = log(x + y)

Use the Product Rule of Logarithm on the left side.

log(xy) = log(x + y)

xy = x + y

Subtract y from both sides.

xy - y = x

Factor.

y(x - 1) = x

Divide both sides by (x - 1).

y = x/(x - 1)

Problem 14 :

Given that

log102 = x

log103 = y

Find the value of log101.2 in terms of x and y.

Solution :

= log101.2

= log10(12/10)

= log1012 - log1010

= log10(4 ⋅ 3) - 1

= log104 + log103 - 1

= log1022 + log103 - 1

= 2log102 + log103 - 1

= 2x + y - 1

Problem 15 :

Solve for x :

100x = log21024 

Solution :

100= log21024 

100= log2210

100= 10log22

100= 10(1)

100= 10

(102)x = 10

102x = 10

102x = 101

2√x = 1

Divide both sides by 2.

√x = 1/2

Square both sides.

(√x)2 = (1/2)2

x = 1/4

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Cross Product Rule in Proportion

    Oct 05, 22 11:41 AM

    Cross Product Rule in Proportion - Concept - Solved Problems

    Read More

  2. Power Rule of Logarithms

    Oct 04, 22 11:08 PM

    Power Rule of Logarithms - Concept - Solved Problems

    Read More

  3. Product Rule of Logarithms

    Oct 04, 22 11:07 PM

    Product Rule of Logarithms - Concept - Solved Problems

    Read More