Given logarithmic equation is `ln(x+30)+lnx=ln31`

We know that `lnx+lny=ln(xy)`

So, `ln((x+30)x)=ln31`

As `lnx=lnyrArrx=y` .

So, `(x+30)x=31`

or, `x^2+30x-31=0`

or, `x^2+31x-x-31=0`

or, `(x+31)(x-1)=0`

or, `x=1` and `x=-31.`

As logarithm for negative numbers is not defined. We can discard x=-31.

So x=1 is the desired solution.

The equation ln (x+30) + ln x = ln 31 has to be solved for x.

ln (x+30) + ln x = ln 31

Use the rule ln x + ln y = ln(x*y)

ln((x+30)*x) = ln 31

=> (x+30)*x = 31

=> x^2 + 30x - 31 = 0

=> x^2 + 31x - x - 31 = 0

=> x(x + 31) - 1(x + 31) = 0

=> (x - 1)(x + 31) = 0

=> x = 1 and x = -31

As the logarithm of a negative number is not defined, eliminate x = -31

**The solution of the equation is x = 1**