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Given

Log (log x) + Log ( Log x³ - 2 ) = 0

=> Log [ (log x) (3 Log x - 2) ] = 0

=> (Log x ) * (3 Log x - 2) = 1

Let Log x = z, So we have

z * (3 z - 2) = 1

3 z² - 2 z - 1 = 0

The factors of 3*-1 are -3 and +1, and their sum is -2.

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

So z = -1/3 or z = 1. we ignore the negative value, for the reason mentioned above.

So Log x = 1

=> x = 10.

Log (log x) + Log ( Log x³ - 2 ) = 0

=> Log [ (log x) (3 Log x - 2) ] = 0

=> (Log x ) * (3 Log x - 2) = 1

*As we have Log (log x) defined, log x is positive.*Let Log x = z, So we have

z * (3 z - 2) = 1

3 z² - 2 z - 1 = 0

The factors of 3*-1 are -3 and +1, and their sum is -2.

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

So z = -1/3 or z = 1. we ignore the negative value, for the reason mentioned above.

So Log x = 1

=> x = 10.