Solve the following using laws of logarithms.
1: log 225
10
2: 3
log 2
2
3: log 1125
Answers
Answer:
Logarithm
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In logarithm we will practice different types of questions on how to solve logarithmic functions on log. Solved examples on logarithm will help us to understand each and every log rules and their applications. Solving logarithmic equation are explained here in details so that student can understand where it is necessary to use logarithm properties like product rule, quotient rule, power rule and base change rule.
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Step-by-step solved example in Log:
1. Find the logarithms of:
(i) 1728 to the base 2√3
Solution:
Let x denote the required logarithm.
Therefore, log2√3 1728 = x
or, (2√3)x = 1728 = 26 ∙ 33 = 26 ∙ (√3)6
or, (2√3)x = (2√3)6
Therefore, x = 6.
(ii) 0.000001 to the base 0.01.
Solution:
Let y be the required logarithm.
Therefore, log0.01 0.000001 = y
or, (0.01y = 0.000001 = (0.01)3
Therefore, y = 3.
2. Proof that, log2 log2 log2 16 = 1.
Solution:
L. H. S. = log2 log2 log2 24
= log2 log2 4 log2 2
= log2 log2 22 [since log2 2 = 1]
= log2 2 log2 2
= 1 ∙ 1
= 1. Proved.
3. If logarithm of 5832 be 6, find the base.
Solution:
Let x be the required base.
Therefore, logx 5832 = 6
or, x6 = 5832 = 36 ∙ 23 = 36 ∙ (√2)6 = (3 √2)6
Therefore, x = 3√2
Therefore, the required base is 3√2
4. If 3 + log10 x = 2 log10 y, find x in terms of y.
Solution:
3 + log10 x = 2 log10 y
or, 3 log10 10 + log10 x= 1og10 y2 [since log10 10 = 1]
or. log10 103 + log10 x = log10 y2
or, log10 (103 ∙ x) = log10 y2
or, 103 x = y2
or, x = y2/1000, which gives x in terms y.
5. Prove that, 7 log (10/9) + 3 log (81/80) = 2log (25/24) + log 2.
Solution:
Since,7 log (10/9) + 3 log (81/80) - 2 log (25/24)
= 7(log 10 – log 9)+ 3(1og 81 - log 80)- 2(1og 25 - 1og 24)
= 7[log(2 ∙ 5) - log32] + 3[1og34 - log(5 ∙ 24)] - 2[log52 - log(3 ∙ 23)]
= 7[log 2 + log 5 – 2 log 3] + 3[4 log 3 - log 5 - 4 log 2] - 2[2 log 5 – log 3 – 3 log 2]
= 7 log 2+ 7 log 5 - 14 log 3 + 12 log 3 – 3 log 5 – 12 log 2 – 4 log 5 + 2 log 3 + 6 log 2
= 13 log 2 – 12 log 2 + 7 log 5 – 7 log 5 – 14 log 3 + 14 log 3 = log 2
Therefore 7 log(10/9) +3 log (81/80) = 2 log (25/24) + log 2. Proved.
6. If log10 2 = 0.30103, log10 3 = 0.47712 and log10 7 = 0.84510, find the values of
(i) log10 45
(ii) log10 105.
(i) log10 45
Solution:
log10 45 = log10 (5 × 9)
= log10 5 + log10 9
= log10 (10/2) + log10 32
= log10 10 - log10 2 + 2 log10 3
= 1 - 0.30103 + 2 × 0.47712
= 1.65321.
(ii) log10 105
Solution:
log10 105
= log10 (7 x 5 x 3)
= log10 7 + log10 5 + log10 3
= log10 7 + log10 10/2 + log10 3
= log10 7 + log10 10 - log10 2 + log10 3
= 0.845l0 + 1 - 0.30103 + 0.47712
= 2.02119.
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