Ineed to find all real solutions. in this problem, a> 0, a is not equal to 1

|log[a] x| = log[a^2] x + 1

log[a] x = log[a^2] x + 1 (Eq. 1)       or      log[a] x = -log[a^2] x - 1 (Eq. 2)

Eq. 1:

(log x)/(log a) = (log x)/(log a^2) + 1

(log x)/(log a) = (log x)/(2log a) + 1

2log x = log x + 2log a

log x = 2log a

log x = log a^2

x = a^2

Eq. 2:

log[a] x = -log[a^2] x - 1

(log x)/(log a) = -(log x)/(log a^2) - 1

(log x)/(log a) = -(log x)/(2log a) - 1

2log x = -log x - 2log a

3log x = -2log a

log x^3 = log a^(-2)

x^3 = a^(-2)

x = a^(-2/3)

x = 1/a^(2/3)

Answers:

x = a^2 or x = 1/a^(2/3)

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