LOGARITHM PROBLEMS WITH SOLUTIONS

Problem 1 :

Find the logarithm of 64 to the base 2√2.

Solution :

Write 64 as in terms of 2√2.

64 = 2⁶

= 2⁴⁺²

= 2⁴ ⋅ 2²

= 2⁴ ⋅ [(√2)²]²

= 2⁴ ⋅ (√2)⁴

= (2√2)⁴

log_2√264 = log_2√2(2√2)⁴

= 4log_2√2(2√2)

= 4(1)

= 4

Problem 2 :

If log_abc = x, log_bca = y and log_cab = z, then find the value of

Solution :

x + 1 = log_abc + log_aa = log_aabc

y + 1 = log_bca + log_bc = log_babc

z + 1 = log_cab + log_cc = log_cabc

Problem 3 :

If a = log₂₄12, b = log₃₆24 and c = log₄₈36, then find the value of (1 + abc) in terms of b and c.

Solution :

1 + abc = 1 + log₂₄12 ⋅ log₃₆24 ⋅ log₄₈36

= 1 + log₃₆12 ⋅ log₄₈36

= 1 + log₄₈12

= log₄₈48 + log₄₈12

= log₄₈(48 ⋅ 12)

= log₄₈(2 ⋅ 12)²

= 2log₄₈24

= 2log₃₆24 ⋅ log₄₈36

= 2bc

Problem 4 :

Find the value of log 0.0001 to the base 0.1.

Solution :

log_0.10.0001 = log_0.1(0.1)⁴

= 4log_0.10.1

= 4(1)

= 4

Problem 5 :

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

Solution :

2logx = 4log3

Divide each side by 2.

logx = 2log3

logx = log3²

logx = log9

x = 9

Problem 6 :

Find the value of log_√264.

Solution :

log_√264 = log_√2(2)⁶

= 6log_√2(2)

= 6log_√2(√2)²

= 6 ⋅ 2log_√2(√2)

= 12 ⋅ 2(1)

= 12

Problem 7 :

Find the value of log (¹⁄₈₁) to the base 9.

Solution :

= log₉(¹⁄₈₁)

= log₉1 - log₉81

= 0 - log₉(9)²

= -2log₉9

= -2(1)

= -2

Problem 8 :

Find the value of log(0.0625) to the base 2.

Solution :

= log₂(0.0625)

= log₂(0.5)⁴

= 4log₂(0.5)

= 4log₂(½)

= 4(log₂1 - log₂2)

= 4(0 - 1)

= 4(-1)

= -4

Problem 9 :

Given log2 = 0.3010 and log3 = 0.4771, find the value of log6.

Solution :

= log6

= log(2 ⋅ 3)

= log2 + log3

Substitute the values of log2 and log3.

= 0.3010 + 0.4771

= 0.7781

Problem 10 :

Find the value of log(0.3) to the base 9.

Solution :

= log₉(0.3)

= log₉(⅓)

= log₉1 - log₉3

= 0 - log₉3

= -log₉3

Problem 11 :

Solve for x :

log₂x - log₂5 = 2 + log₂3

Solution :

log₂x - log₂5 = 2 + log₂3

Use the fundamental law of logarithm.

log₂(ˣ⁄₅) = 2 + log₂3

Subtract log₂3 from each side.

log₂(ˣ⁄₅) - log₂3 = 2

log₂(ˣ⁄₁₅) = 2

Convert to exponential form.

ˣ⁄₁₅ = 2²

ˣ⁄₁₅ = 4

Multiply each side by 15.

x = 60

Problem 12 :

Find integer values of m and n for which 

m - nlog₃2 = 10log₉6

Solution :

m - nlog₃2 = 10log₉6

Use the property of logarithm.

m - nlog₃2 = 5log₃6

m - nlog₃2 = 5log₃(3 ⋅ 2)

Use the fundamental law of logarithm.

m - nlog₃2 = 5[log₃3 + log₃2]

m - nlog₃2 = 5[1 + log₃2]

Use the distributive property.

m - nlog₃2 = 5 + 5log₃2

Equate the constant terms and coefficients of like terms.

Problem 13 :

Solve the simultaneous equations :

log₂6x = 1 + 2log₂y

1 + log₆x = log₆(15y - 25)

Solution :

log₂6x = 1 + 2log₂y ----(1)

1 + log₆x = log₆(15y - 25) ----(2)

(1) :

log₂6x = 1 + 2log₂y

log₂6x = log₂2 + 2log₂y

log₂6x = log₂2 + log₂y²

log₂6x = log₂(2y²)

6x = 2y² ----(3)

(2) :

1 + log₆x = log₆(15y - 25)

log₆6 + log₆x = log₆(15y - 25) 

log₆(6x) = log₆(15y - 25) 

6x = 15y - 25

From (3), substitute 2y² for 6x in (4).

2y² = 15y - 25

2y² - 15y + 25 = 0

Factor and solve.

2y² - 10y - 5y + 25 = 0

2y(y - 5) - 5(y - 5) = 0

(y - 5)(2y - 5) = 0

y = 5  or  y = ⁵⁄₂

Substitute 5 and ⁵⁄₂ for y in (3).

Therefore, 

x = ²⁵⁄₃, y = 5

x = ²⁵⁄₁₂, y = ⁵⁄₂

Problem 14 :

Solve for x :

log₂(x + 3) + log₂(x - 3) = 4

Solution :

log₂(x + 3) + log₂(x - 3) = 4

Use the fundamental law of logarithm.

log₂[(x + 3)(x - 3)] = 4

Convert to exponential form.

(x + 3)(x - 3) = 2⁴

x² - 3² = 16

x² - 9 = 16

x² = 25

√x² = √25

x = ± 5

x = -5 or 5

But, x = -5 will not work with the given equation.

Because, x = -5 makes the arguments negative in the logarithms. Since logarithm is defined only for positive value in argument, x = -5 is not a solution.

So, x = 5 is the solution to the above equation.

Problem 15 :

Prove that :

Solution :

Let

We can take the expression in (1) and prove, that is equivalent to the expression in (2).

So, (1) = (2). 

Hence, proved.

Video Lessons

Introduction to Logarithms

Fundamental Laws of Logarithms

Change of Base

Using Log Table

Using Antilog Table

Important Stuff

Difference between Bar Value and Negative Value

Multiplication of Two Logarithms

Relationship between Exponents and Logarithms

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Problem - 2

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Problem - 9

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Problem - 22

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