How do you solve $log_4 (x + 4) - log_4 (x - 4) = log_4 3$ ?

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Answer 1 · 2015-12-06T10:23:12

$x=8$

To solve this kind of equations, the strategy is to manipulate the expressions in order to arrive at something like

$log(X) = log(Y)$

To deduce that $X=Y$ , since the logarithm function is injective.

So, the only thing we need to do in this case it to remember that

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

So you have that

$log_4((x+4)/(x-4)) = log_4(3)$

Which is true if and only if

$(x+4)/(x-4) = 3$

If $x \ne 4$ , we can multiply by $x-4$ both terms and get

$x+4 = 3x-12 \to 2x=16 \to x=8$

How do you solve log_4 (x + 4) - log_4 (x - 4) = log_4 3log_4 (x + 4) - log_4 (x - 4) = log_4 3?