Question

How do you solve `log x + log (x+1) = 1`?

Answer 1

Derive a quadratic, one of whose roots is an acceptable solution of the equation, namely:

`x = (sqrt(41)-1)/2`

Explanation:

From the basic properties of logs we have:

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

Since `log` is a one-one function from `(0,oo) -> RR`, we require:

`10 = x(x+1)`

That is:

`x^2+x-10 = 0`

This is of the form `ax^2+bx+c` with `a=1`, `b=1` and `c=-10`

Use the quadratic formula to find:

`x = (-b+-sqrt(b^2-4ac))/(2a) = (-1+-sqrt(41))/2`

Since `sqrt(41) > 1` only one of these roots is positive, namely:

`x = (-1+sqrt(41))/2 = (sqrt(41)-1)/2`

We can discard the other root of the quadratic since we are dealing with `log` as a function of non-negative Real numbers, so we disallow `log` of negative numbers.