You may use the following method, based on property of logarithmic function such that:

`log(x^2 + 7x) = 1 => log(x^2 + 7x) = log 10 (1 is replaced by log 10)`

Since the logarithmic function is bijective, yields:

`x^2 + 7x = 10 `

You need to complete the square to the left side such that:

`x^2 + 2*(7/2)x + (7/2)^2 = 10 + (7/2)^2`

`(x + 7/2)^2 = (40+49)/4 => (x + 7/2)^2 = 89/4`

`(x + 7/2) = +- sqrt89/2 => x_1 = sqrt89/2 - 7/2`

`x_2 = -(sqrt89/2 + 7/2)`

Since the logarithms are defined for `x>0` , hence, the negative value of x, `x_2 = -(sqrt89/2 + 7/2)`  does not satisfy the condition `x>0` .

Since `sqrt 89 > sqrt81 = 9`  yields:

`sqrt89/2 - 7/2 > sqrt81/2 - 7/2 = 9/2 - 7/2 = 1 > 0 `

Hence, evaluating the solution to the given equation yields `x_1 = sqrt89/2 - 7/2` .

The equation `log(x^2 + 7x) = 1` has to be solved for x.

`log(x^2 + 7x) = 1`

log represents logarithm to the base 10

=> `x^2 + 7x = 10^1`

=> `x^2 + 7x = 10`

=> `x^2 + 7x - 10 = 0`

The solutions of this quadratic equations are

`(-7 + sqrt(49 + 40))/2 = (-7 + sqrt 89)/2` and `(-7 - sqrt 89)/2`

As the logarithm of negative numbers is not defined, take only the solution `x = (-7 + sqrt 89)/2`

The solution of the equation is `(-7 + sqrt 89)/2`