1/3log(base5) 3x + log(base5) 4=3
what I meant is: 1/3=one-third ,log5= log(base5)
as I noted in your earlier post,
assuming base 5 throughout,
1/3 log3x + log4 = 3
log3x + 3log4 = 9
log(3x * 4^3) = 9
192x = 5^9
x = 5^9/192
what I meant is: 1/3=one-third ,log5= log(base5)
assuming base 5 throughout,
1/3 log3x + log4 = 3
log3x + 3log4 = 9
log(3x * 4^3) = 9
192x = 5^9
x = 5^9/192