How do you solve $ln x + ln (x + 6) = \frac{1}{2} ln 9$ $ln x + ln (x+6) = 1/2 ln 9$ ?

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Answer 1 · 2015-06-20T01:25:04

$x = - 3 + \sqrt{12}$ $x=-3+\sqrt{12}$
Use rules for adding logarithms and rule for bringing the coefficient inside the logarithm as an exponent.

$ln (A) + ln (B) = ln (A B)$ $ln(A)+ln(B)=ln(AB)$

so

$ln (x) + ln (x + 6) = ln (x^{2} + 6 x)$ $ln(x)+ln(x+6)=ln(x^2+6x)$

$C ln (D) = ln (D^{C})$ $Cln(D)=ln(D^C)$

so

$\frac{1}{2} ln (9) = ln (9^{\frac{1}{2}}) = ln (3)$ $1/2ln(9)=ln(9^{1/2})=ln(3)$

Making these substitutions,

$ln (x) + ln (x + 6) = \frac{1}{2} ln (9)$ $ln(x)+ln(x+6)=1/2ln(9)$

becomes

$ln (x^{2} + 6 x) = ln (3)$ $ln(x^2+6x)=ln(3)$

which requires,

$x^{2} + 6 x = 3$ $x^2+6x=3$

$\Rightarrow x^{2} + 6 x - 3 = 0$ $\implies x^2+6x-3=0$

$x = \frac{- 6 \pm \sqrt{6^{2} - 4 (1) (- 3)}}{2 (1)}$ $x={-6\pm\sqrt{6^2-4(1)(-3)}}/{2(1)}$

$x = - 3 \pm \sqrt{12}$ $x=-3\pm\sqrt{12}$

$ln (x)$ $ln(x)$ requires $x > 0$ $x>0$ so we choose the plus sign

$x = - 3 + \sqrt{12}$ $x=-3+sqrt{12}$

How do you solve ln x + ln (x+6) = 1/2 ln 9lnx+ln(x+6)=12ln9ln x + ln (x+6) = 1/2 ln 9?