find log_9(square root of 3)?

Answer 1

let log = log to base 9:-
log 3^(1/2) = (1/2) log 3 = (1/2)(1/2) = 1/4

Answer 2

log_9(3^(1/2)) = log_9(9^(1/4)) = 1/4
since 9^(1/2) = 3, we can raise both sices to the 1/2 again and get:
9^(1/4) = 3^(1/2) and we plug that in to the equation to get the answer. Once we have the same base we can just cancel the bases to get the answer 1/4 as required.

Answer 3

log_9(square root of 3)?

9^x=√3 (ln both sides)
xln9=ln√3
x=1/4

Answer 4

log_9(sqrt3) = x

9^x = sqrt 3
9^x = 3^(1/2)
3^(2x) = 3^(1/2)

2x = 1/2
x = 1/4

log_9 (sqrt3) = 1/4

Answer 5

if y = log9[sqrt(3)] then
9^y = sqrt(3), and since the sqrt(3) = 3^(1/2) and 3 = 9^(1/2) then
9^y = [9^(1/2)]^(1/2) = 9^(1/4), if b^m = b^n, then m = n
y = 1/4, substitute y for log9[sqrt(3)]
log9[sqrt(3)] = 1/4

Answer 6

log_9 sqrt(3)= log_9(3^(1/2))
x=(1/2)log_9(3)
2x=log_9(3)
9^2x=3
2x (log(9))=log(3)
2x=log(3)/log(9)
2x=log(3)/log(3^2)
2x=log(3)/2log(3)
2x=1/2
x=1/4

Answer 7

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