Question

solve a) log 9 base x = 2/3
b) log x base 9 = 0.5
c) log x base 4 = 2
d) [log (x + 1) base 5]^2 = 4
e) log 3 base 4 + log (x+2) base 4 = 2​

Answer 1

Answer:

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Step-by-step explanation:

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 -