Question

How do you solve `log_[2] (x+20) - log_[2] (x+2) = log_[2] x`?

Answer 1

`x = 4`

Explanation:

First of all, let's establish the domain. As any logarithmic expression can only have positive arguments, the following conditions must hold:

• In order for the first logarithmic expression to be defined, `x + 20 > 0 " "<=>" " x > -20` must hold.
• For the second logarithmic expression, `x + 2 > 0 <=> x > -2` must hold.
• For the third logarithmus, `x > 0` must hold.

As `x > 0` is the strictest condition of the three (all three conditions would be satisfied if this one is satisfied).

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Now, let's solve the equation.

First of all, we can use the logarithmic law

`log_a (m) - log_a(n) = log_a (m/n)`

In this case, we get:

`log_2(x+20) - log_2(x+2) = log_2(x)`

`<=> log_2 ((x+20)/(x+2)) = log_2(x)`

Now that on both sides we have just one logarithmic expression, we can use `log x = log y` if and only if `x = y` and drop the logarithms:

`<=> (x+20)/(x+2) = x`

... multiply both sides with `(x +2)`...

`<=> x + 20 = x(x+2)`

`<=> x^2 + x -20 = 0`

The solutions for this quadratic equation are `x = -5` or `x = 4`.

However, as we have established that `x > 0` must hold in order for our logarithmic expressions to be defined, `x = -5` can't be a solution.

Thus, our only solution is `x = 4`.

Answer 2

I found `x=4`

Explanation:

Try this: