Can you help me with this logarithm? (alg2)

2logx-log2x= 0.301


and


log(base of 3)√(2x-1) = 3

Question 1)
... 2log(x) - log(2x) = 0.301
or log(x^2) - log(2x) = log( 10^0.301 )
or x^2 /(2x) = 10^0.301
or x = 2*10^0.301
or x ~ 4

Question 2)
... log3(√(2x-1)) = 3
or √(2x-1) = 3^3
or 2x - 1 = 27^2
or 2x = 1+ 729
or x = 365