If `log_x(1/4) = -2/3` what is x

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x has to be determined given that `log_x(1/4) = -2/3 `

`log_x(1/4) = -2/3`

=> `(1/4) = x^(-2/3)`

=> `(1/8)^(2/3) = x^(-2/3)`

=> `8^(-2/3) = x^(-2/3)`

As the base is the same equate the exponent.

=> 8 = x

**The solution of the equation is x = 8**

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