Problem 1 :
Find the logarithm of 64 to the base 2√2.
Solution :
Write 64 as in terms of 2√2.
64 = 2^{6}
= 2^{4+2}
= 2^{4 }⋅ 2^{2}
= 2^{4 }⋅ [(√2)^{2}]^{2}
= 2^{4 }⋅ (√2)^{4}
= (2√2)^{4}
log_{2√2}64 = log_{2√2}(2√2)^{4}
= 4log_{2√2}(2√2)
= 4(1)
= 4
Problem 2 :
If log_{a}bc = x, log_{b}ca = y and log_{c}ab = z, then find the value of
1/(x + 1) + 1/(y + 1) + 1/(z + 1)
Solution :
x + 1 = log_{a}bc + log_{a}a = log_{a}abc
y + 1 = log_{b}ca + log_{b}c = log_{b}abc
z + 1 = log_{c}ab + log_{c}c = log_{c}abc
1/(x + 1) = 1/log_{a}abc = log_{abc}a
1/(y + 1) = 1/log_{b}abc = log_{abc}b
1/(z + 1) = 1/log_{c}abc = log_{abc}c
1/(x + 1) + 1/(y + 1) + 1/(z + 1) = log_{abc}a + log_{abc}b + log_{abc}c
= log_{abc}abc
= 1
Problem 3 :
If a = log_{24}12, b = log_{36}24 and c = log_{48}36, then find the value of (1 + abc) in terms of b and c.
Solution :
1 + abc = 1 + log_{24}12 ⋅ log_{36}24 ⋅ log_{48}36
= 1 + log_{36}12 ⋅ log_{48}36
= 1 + log_{48}12
= log_{48}48 + log_{48}12
= log_{48}(48 ⋅ 12)
= log_{48}(2 ⋅ 12)^{2}
= 2log_{48}24
= 2log_{36}24 ⋅ log_{48}36
= 2bc
Problem 4 :
Find the value of log 0.0001 to the base 0.1.
Solution :
log_{0.1}0.0001 = log_{0.1}(0.1)^{4}
= 4log_{0.1}0.1
= 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 = log3^{2}
logx = log9
x = 9
Problem 6 :
Find the value of log_{√2}64.
Solution :
log_{√2}64 = 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 :
log_{9}(1/81) = log_{9}1 - log_{9}81
= 0 - log_{9}(9)^{2}
= -2log_{9}9
= -2(1)
= -2
Problem 8 :
Find the value of log(0.0625) to the base 2.
Solution :
log_{2}(0.0625) = log_{2}(0.5)^{4}
= 4log_{2}(0.5)
= 4log_{2}(1/2)
= 4(log_{2}1 - log_{2}2)
= 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 :
log_{9}(0.3) = log_{9}(1/3)
= log_{9}1 - log_{9}3
= 0 - log_{9}3
= -log_{9}3
= -1 / log_{3}9
= -1 / log_{3}3^{2}
= -1 / 2log_{3}3
= -1 / 2(1)
= -1/2
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