How do you solve $log (x+4)=log x + log 4$ ?

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Answer 1 · 2015-08-15T15:40:49

$log(x+4) = log(x) + log(4) = log(4x)$ , so $x+4 = 4x$ , hence $x = 4/3$

If $a, b > 0$ then $log(a)$ and $log(b)$ are defined and

$log(a) + log(b) = log(ab)$

So $log(x) + log(4) = log(4x)$

So our equation becomes:

$log(x + 4) = log x + log 4 = log(4x)$

Since $log:(0, oo) -> RR$ is one-one, that means:

$x+4 = 4x$

Subtract $x$ from both sides to get $4 = 3x$ , then divide both sides by $3$ to get $x = 4/3$ .

How do you solve log (x+4)=log x + log 4log (x+4)=log x + log 4?