Question

How do you solve `log (x + 3) = log (6) - log (2x-1)`?

Answer 1

`x = (-5 + sqrt(97))/4`

Explanation:

I'm assuming that all the `log` functions have the same basis `> 1`.

First of all, let's compute the domain of the logarithmic expressions.
For the first one, `x + 3 > 0` must hold, so `x > -3`.
For the last one, we get `2x -1 > 0 <=> x > 1/2`.

As the second condition is the more restrictive one, we can assume that the domain is `x > 1/2` and any possible solutions need to respect this condition.

Now, in order to "get rid" of the logarithmic expressions, it is necessary to simplify the terms on the right-hand side.

To do so, remember the logarithmic rule: `log_a(x) - log_a(y) = log_a(x/y)`

In our case, it means:

`log(x+3) = log(6) - log(2x-1)`
`<=> log(x+3) = log(6/(2x-1))`

Now, due to the fact that for `x`, `y > 0` and `a != 1`, `log x = log y <=> x = y` holds, we can "drop" the logarithms on both sides of the equation.

`<=> x + 3 = 6/(2x-1)`

... multiply both sides with `2x-1`...

`<=> (x+3)(2x-1) = 6`

`<=> 2x^2 +5x - 9 = 0`

This is a quadratic equation that can be solved e. g. with the quadratic formula:

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

Here, `a = 2`, `b = 5`, `c = -9`, so we can compute the solution as follows:

`x = (-5 +- sqrt(25 - 4 * 2 * (-9))) / 4 = (-5 +- sqrt(97))/4`

The solution `x = (-5 - sqrt(97))/4` is negative, therefore we can ignore it because of our restriction to the domain `x > 1/2`.

So, we have just one solution: `x = (-5 + sqrt(97))/4 ~~ 1.2122`

Hope that this helped!