Let x DENOTE the required logarithm.Therefore, log2√3 1728 = xor, (2√3)x = 1728 = 26 ∙ 33 = 26 ∙ (√3)6or, (2√3)x = (2√3)6Therefore, x = 6.(ii) 0.000001 to the base 0.01.Solution:Let y be the required logarithm.Therefore, log0.01 0.000001 = yor, (0.01y = 0.000001 = (0.01)3Therefore, 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 = 6or, X6 = 5832 = 36 ∙ 23 = 36 ∙ (√2)6 = (3 √2)6Therefore, x = 3√2Therefore, the required base is 3√24. If 3 + log10 x = 2 log10 y, find x in terms of y.Solution:3 + log10 x = 2 log10 yor, 3 log10 10 + log10 x= 1og10 y2 [since log10 10 = 1]or. log10 103 + log10 x = log10 y2or, log10 (103 ∙ x) = log10 y2or, 103 x = y2or, 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 2Step-by-step explanation:hope you will understand ☺️