# LOGARITHM PROBLEMS

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

1/(x + 1) + 1/(y + 1) + 1/(z + 1)

Solution :

x + 1 = logabc + logaa = logaabc

y + 1 = logbca + logbc = logbabc

z + 1 = logcab + logcc = logcabc

1/(x + 1) = 1/logaabc = logabca

1/(y + 1) = 1/logbabc = logabcb

1/(z + 1) = 1/logcabc = logabcc

1/(x + 1) + 1/(y + 1) + 1/(z + 1) = logabca + logabcb + logabcc

= logabcabc

= 1

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(1/2)

= 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(1/3)

= log91 - log93

= 0 - log93

= -log93

= -1 / log39

= -1 / log332

= -1 / 2log33

= -1 / 2(1)

= -1/2

## 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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