How do you solve $log_2 (x - 1) – log_2 (x – 4) = log_2 4$ ?

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Answer 1 · 2015-12-01T11:19:30

$x=5$

We have that $log(a)-log(b)=log(a/b)$ , so

$log_2(x-1)-log_2(x-4) = log_2((x-1)/(x-4))$

The equation has thus become

$log_2((x-1)/(x-4))=log_2(4)$

This means that

$(x-1)/(x-4) = 4$

If $x \ne 4$ , multiply both sides by $x-4$ , obtaining

$x-1 = 4x-16$ , and solve as usual

$-3x = -15 \to x=5$

Verify the answer:

If $x=5$ , the expression becomes

$log_2(5-1) - log_2(5-4) = log_2(4)$

$log_2(4) - log_2(1) = log_2(4)$

which is true, since $log_a(1)=0$ for every $a$ .

How do you solve log_2 (x - 1) – log_2 (x – 4) = log_2 4log_2 (x - 1) – log_2 (x – 4) = log_2 4?