How do you solve $log 3 x = log 4 + log (x + 3)$ $Log3x=log4+log(x+3)$ ?

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Answer 1 · 2016-04-19T04:28:04

x > 0. The negative solution $x = - 12$ $x = -12$ is inadmissible for the given equation. So there is no real solution.

x > 0.
Use $log a - log b = log (\frac{a}{b})$ $log a - log b = log (a/b)$

$log (\frac{3 x}{x + 3}) = log 4$ $log ((3x)/(x+3))=log 4$

Som $\frac{3 x}{x + 3} = 4$ $(3x)/(x+3)=4$

$x = - 12$ $x = -12$ .

As x > 0 for log x, $x = - 12$ $x = -12$ is inadmissible.

How do you solve Log3x=log4+log(x+3)log3x=log4+log(x+3)Log3x=log4+log(x+3)?