# How do you solve log_ (x+20) - log_ (x+2) = log_ x?

Feb 4, 2016

$x = 4$

#### 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, $x + 20 > 0 \text{ "<=>" } x > - 20$ must hold.
• For the second logarithmic expression, $x + 2 > 0 \iff x > - 2$ 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

${\log}_{a} \left(m\right) - {\log}_{a} \left(n\right) = {\log}_{a} \left(\frac{m}{n}\right)$

In this case, we get:

${\log}_{2} \left(x + 20\right) - {\log}_{2} \left(x + 2\right) = {\log}_{2} \left(x\right)$

$\iff {\log}_{2} \left(\frac{x + 20}{x + 2}\right) = {\log}_{2} \left(x\right)$

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:

$\iff \frac{x + 20}{x + 2} = x$

... multiply both sides with $\left(x + 2\right)$...

$\iff x + 20 = x \left(x + 2\right)$

$\iff {x}^{2} + x - 20 = 0$

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 $x = 4$.

Feb 4, 2016

I found $x = 4$

#### Explanation:

Try this: 