You need to convert 1 into `log 10` and `2log3` into `log 3^2` such that:

`log 10 + log 9 - log x = log(x + 1)`

You need to use the properties of logarithms to convert the sum `log 10 + log 9` into the logarithm of product, such that:

`log 10 + log 9 = log(10*9) => log 10 + log 9 = log 90`

Substituting `log 90` for `log 10 + log 9` yields:

`log 90 - log x = log(x + 1)`

Using again the properties of logarithms yields:

`log (90/x) = log(x + 1) => (90/x) = (x + 1)`

`90 = x^2 + x => x^2 + x - 90 = 0`

You may use factorization to solve quadratic equation such that:

`x^2 + x - 81 - 9 = 0 => (x^2 - 81) + x - 9 = 0`

`(x - 9)(x + 9) + (x - 9) = 0`

Factoring out `(x - 9)` yields:

`(x - 9)(x + 9 + 1) = 0 => {(x - 9 = 0),(x + 10 = 0):} => {(x = 9),(x = -10):}`

You need to test the values -10 and 9 in equation, such that:

`log90 - log(-10) = log(-10+1)`

Since `log-10` and `log-9` are invalid values, hence, the solution x = -10 is invalid.

`log90 - log9 = log(9+1) => log(90/9) = log10 => log10 = log10`

**Hence, evaluating the solution to the given equation yields **`x = 9.`

First, we'll impose the constraints of existence of logarithms:

x>0

and

x+1>0

x>-1

The common interval of admissible values for x is (0 , +inf.).

We'll re-write the term 2log 3 using the power rule of logarithms:

2log 3 = log 3^2

2log 3 = log 9

We'll re-write the equation, moving all terms that contain the variable x to one side:

1 + log 9 = log(x+1) + log x

Now, we'll solve the equation, applying th product rule of the logarithms:

log9 + log 10 = lg(x+1) + lgx

log 9*10 = log[x*(x+1)]

Because the bases of logarithms are matching, we'll use the one to one property:

x*(x+1)=9*10

We'll remove the brackets and we'll move all terms to one side:

x^2 +x -90=0

We'll apply the quadratic formula:

x1=[-1+ sqrt(1+4*90)]/2=(-1+19)/2=9 > 0

x2=(-1-19)/2=-10 < 0

We'll reject the second solution and the equation will have just the solution x = 9.