`6^(log^2_6 x) + x^(log_6 x)=12`

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You should use the following logarithmic identity such that:

`a^(log_a b) = b`

`a^(x^2) = (a^x)^x`

`6^(log^2_6 x) = (6^(log_6 x))^(log_6 x) = x^(log_6 x)`

Hence, substituting `x^(log_6 x)` for `6^(log^2_6 x)` yields:

`x^(log_6 x) + x^(log_6 x) = 12`

`2x^(log_6 x) = 12 => x^(log_6 x) = 12/2 => x^(log_6 x) = 6`

You should take logarithms both sides such that:

`log_6 (x^(log_6 x)) = log_6 6`

`(log_6 x)(log_6 x) = 1 => log^2_6 x = 1 => log_6 x = +-1`

`x_1 = 6 ; x_2 = 6^(-1) => x_2 = 1/6`

**Hence, evaluating the solutions to the given equation yields `x_1 = 6` and `x_2 = 1/6` .**

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