You need to rewrite the term `lg(x^5)` using the power property of logarithms, such that:

`lg^2 x + 5lg x - 6 = 0`

You should come up with the notation ` lg x = y` , such that:

`y^2 + 5y - 6 = 0`

You need to use quadratic formula, such that:

`y_(1,2) = (-5+-sqrt(25 + 24))/2 => y_(1,2) = (-5+-sqrt49)/2`

`y_1 = 1; y_2 = -6`

You need to solve for `x` the following equations, such that:

`lg x = 1 => x = 10^1 => x = 10`

`lg x = -6 => x = 10^(-6) => x = 1/10^6 `

**Hence, evaluating the solutions to the given equation, yields **`x = 1/10^6 , x = 10.`

`(log x)^2 +log(x^5)-6=0`

`(log x)^2 +5log(x)-6=0` `(log x^n = n*(log x)`

Let u = log x. Then

`u^2+5u-6=0`

`(u+6)(u-1) = 0`

`u = -6, u = 1`

Since u = log x and log x > 0, then u = -6 is extraneous.

Thus, u = log x = 1, or x = 10^1 = 10.