How do you solve $ln x + ln 3 x = 12$ $ln x + ln 3x = 12$ ?

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Answer 1 · 2015-12-06T14:58:20

$x = + \frac{\sqrt{3}}{3} e^{6}$ $x = + sqrt(3)/3 e^6$

$ln (x \cdot 3 x) = 12$ $ln (x * 3x) = 12$

$x \cdot 3 x = e^{12}$ $x * 3x = e^12$

$x^{2} = \frac{1}{3} e^{12}$ $x^2 = 1/3 e^12$

$x = \pm \frac{\sqrt{3}}{3} e^{6}$ $x = ± sqrt(3)/3 e^6$

But x > 0.

How do you solve lnx+ln3x=12ln x + ln 3x = 12?