i will use the following notation: log[a]b means log to the base a of b

log[3]y = (1/2)log[3]81

=> log[3]y = log[3]81^(1/2) ...........since nlogx = log(x^n)

=> log[3]y = log[3]9

=> y = 9

log[6](x + 1) - log[6] (x - 1) = log[6]9

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Problem 2

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log6 (x+1) - log6 (x-1) = log6 9

=> log[6]{(x + 1)/(x - 1)} = log[6]9 ............since logx - logy = log(x/y)

=> (x + 1)/(x - 1) = 9

=> x + 1 = 9x - 9

=> 8x = 10

=> x = 5/4

log[4]n = 2log[4]8

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Problem 3

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log4 n = 2 log4 8

=> log[4]n = log[4]8^2 ..............since nlogx = log(x^n)

=> log[4]n = log[4]64

=> n = 64

is it to evaluate log[4] (7/3)? or is it log[10]?

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Problem 4

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Use log4 7 = 1.4037 and log4 3 = 0.7925 to evaluate log 7/3