How do you solve $log (x + 2) - log (x - 2) = log (3)$ $Log(x+2) - log(x-2) = log(3)$ ?

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Answer 1 · 2015-09-06T17:02:34

$x = 4$ $x=4$

First we simplify the left side by using the rule: $log (a) - log (b) = log (\frac{a}{b})$ $log (a) - log(b) = log(a/b)$

$log (x + 2) - log (x - 2) = log (3)$ $log(x+2) - log(x-2) = log(3)$

$log (\frac{x + 2}{x - 2}) = log (3)$ $log((x+2)/(x-2)) = log(3)$

Then, we apply the rule: $log (a) = b \Leftrightarrow 10^{b} = a$ $log(a) = b <=> 10^b = a$

$10^{log (3)} = \frac{x + 2}{x - 2}$ $10^log(3) = (x+2)/(x-2)$

Then, we can apply the rule: $10^{log (k)} = k$ $10^log(k) = k$

$3 = \frac{x + 2}{x - 2}$ $3 = (x+2)/(x-2)$

Now we can just solve the equation:

$3 x - 6 = x + 2$ $3x-6 = x+2$

$2 x = 8$ $2x=8$

$x = 4$ $x=4$

Alternatively, we can skip a lot of steps by taking a shortcut:

Since $log ()$ $log()$ is a one-to-one function, $log (a) = log (b)$ $log(a) = log(b)$ means $a = b$ $a=b$ .

$log (x + 2) - log (x - 2) = log (3)$ $log(x+2) - log(x-2) = log(3)$

$log (\frac{x + 2}{x - 2}) = log (3)$ $log((x+2)/(x-2)) = log(3)$

$\frac{x + 2}{x - 2} = 3$ $(x+2)/(x-2) = 3$

$x + 2 = 3 x - 6$ $x+2=3x-6$

$8 = 2 x$ $8=2x$

$4 = x$ $4=x$

How do you solve log(x+2)−log(x−2)=log(3)Log(x+2) - log(x-2) = log(3)?