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
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
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