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

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Answer 1 · 2016-02-04T17:12:47

$x = 4$

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 · 2016-02-04T17:28:05

I found $x=4$

Try this:
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How do you solve log_[2] (x+20) - log_[2] (x+2) = log_[2] xlog_[2] (x+20) - log_[2] (x+2) = log_[2] x?