# 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

= 2 22

= 2⋅ [(√2)2]2

= 2⋅ (√2)4

= (2√2)4

log2√264 = log2√2(2√2)4

= 4log2√2(2√2)

= 4(1)

= 4

Problem 2 :

If logabc = x, logbca = y and logcab = z, then find the value of

Solution :

x + 1 = logabc + logaa = logaabc

y + 1 = logbca + logbc = logbabc

z + 1 = logcab + logcc = logcabc

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 :

log0.10.0001 = log0.1(0.1)4

= 4log0.10.1

= 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()

= 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.

 m = 5 -n = 5n = -5

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).

 6x = 2(5)26x = 2(25)6x = 50x = ⁵⁰⁄₆x = ²⁵⁄₃ 6x = 2(⁵⁄₂)26x = 2(²⁵⁄₄)6x = ²⁵⁄₂x = ²⁵⁄₁₂

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

Problem - 1

Problem - 2

Problem - 3

Problem - 4

Problem - 5

Problem - 6

Problem - 7

Problem - 8

Problem - 9

Problem - 10

Problem - 11

Problem - 12

Problem - 13

Problem - 14

Problem - 15

Problem - 16

Problem - 17

Problem - 18

Problem - 19

Problem - 20

Problem - 21

Problem - 22

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