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

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Answer 1 · 2015-06-20T01:25:04

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

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

so

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

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

so

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

Making these substitutions,

$ln(x)+ln(x+6)=1/2ln(9)$

becomes

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

which requires,

$x^2+6x=3$

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

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

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

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

$x=-3+sqrt{12}$

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