Solve for x (logarithms)

Question

3 ln x = log_x e - 2

Answer 1

3lnx = log_xe - 2

3lnx = lne / lnx - 2

3lnx = 1 / lnx - 2

3(lnx)² + 2(lnx) - 1 = 0

Let u = lnx. Then, 3u² + 2u - 1 = 0

(3u - 1)(u + 1) = 0

u = 1/3 or u = -1

So, lnx = 1/3 or lnx = -1

x = e^1/3 or 1/e