Question

How do you solve `log(x^2+4)-log(x+2)=2+log(x-2)`?

Answer 1

`x = sqrt(404/99) ~~2.02...`

Explanation:

1) Domain of the logarithmic expressions

The first thing you need to do is establish the domain of your `log` expressions.

• `log(x^2 + 4)` is defined for `x^2 + 4 > 0` which is true for all `x in RR`
• `log(x+2)` is defined for `x + 2 > 0 <=> x > -2`
• `log(x-2)` is defined for `x - 2 > 0 <=> x > 2`

So, in total, the most restrictive condition is `x> 2` which is our domain.

2) Transform the equation and unite the logarithmic terms

Now, to "get rid" of the logarithmic terms, first of all, you need to eliminate the sums and unite your logarithmic terms.
The goal is to have just one `log` expression (or none at all) on each side of the equation.

Use the logarithmic laws:

`log(a) + log(b) = log(a*b)`

`log(a) - log(b) = log(a/b)`

Now you can transform your equation as follows:

`log(x^2+4 ) - log(x+2) = 2 + log(x-2)`

... subtract `log(x-2)` on both sides...

`<=> log(x^2+4 ) - log(x+2) - log(x-2) = 2`

... use the logarithmic laws...

`<=> log((x^2+4)/((x+2)(x-2))) = 2`

`<=> log((x^2+4)/(x^2 - 4)) = 2`

3) Eliminate the logarithmus

Now, you can "get rid" of the `log`. As you haven't specified the basis, I will assume that the basis of your logarithm is `10`.

The inverse function of `log_10(x)` is `10^x` which means that both `log_10(10^x) = x` and `10^(log_10(x)) = x` hold.

So, to eliminate the `log_10`, you need to apply the function `10^x` to both sides of the equation.

`<=> 10^(log_10((x^2+4)/(x^2 - 4))) = 10^2`

`<=> (x^2+4)/(x^2 - 4) = 10^2`

4) Solve the quadratic equation

`(x^2+4)/(x^2 - 4) = 10^2`

`<=> x^2 + 4 = 100(x^2-4)`

`<=> -99 x^2 = -404`

`<=> x^2 = 404/99`

`<=> x = +- sqrt(404/99) ~~ +- 2.02...`

5) Determine the solution w.r.t. the domain

As we have established earlier that our domain is `x> 2`, only one of the two possible solutions fits the domain.

Thus, we need to discard the negative solution, and the only solution in `RR` is `x = sqrt(404/99) ~~2.02...`