Question

Show that, 7 log (15/16)+ log (8/3) +5 log 2/5 +log (32/25)= log3

Answer 1

Answer:

Step-by-step explanation:

True

According to rule=>

Log A + Log B = Log(A.B)

So,

Log 3 + log 5

Log (3. 5)

Log 15

Answer 2

Answer:

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