How do you solve $ln 3 x + ln 2 x = 3$ $ln 3x+ ln 2x=3$ ?

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Answer 1 · 2015-12-14T08:56:26

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

Use the fact that $ln (a b) = ln (a) + ln (b)$ $ln(ab)=ln(a)+ln(b)$ for positive $a$ $a$ and $b$ $b$ .

$ln 3 x + ln 2 x = (ln 3 + ln x) + (ln 2 + ln x)$ $ln3x+ln2x=(ln3+lnx)+(ln2+lnx)$

$= ln 6 + 2 ln x$ $=ln6+2lnx$

$= 3$ $=3$

Perform algebraic manipulation yields

$ln x = \frac{3 - ln 6}{2}$ $lnx=(3-ln6)/2$

$x = e^{\frac{3 - ln 6}{2}}$ $x=e^(frac{3-ln6}{2})$

$= \frac{e^{\frac{3}{2}}}{e^{\frac{ln 6}{2}}}$ $=frac{e^(frac{3}{2})}{e^(frac{ln6}{2})}$

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

How do you solve ln3x+ln2x=3ln 3x+ ln 2x=3?