Solving this equation: 2logbase3(x) = logbase3(81)

O K, we are working in base 3. I will assume the 3 is there, and not repeat it endlessly.

2 log x = log 81

log x^2 = log 3^4 . . . . . . . (because 3^4 = 81)

Therefore

x^2 = 3^4

√(x^2) = √(3^4)

x = 3^2

x = 9

2 log₃(x) = log₃(81)
log₃(x^2) = log₃(81)

x^2 = 81
x = -9, 9

However, log can only take positive arguments, so we need x > 0

x = 9

X=9