How do you solve $ln (x - 6) + ln (x + 3) = ln 22$ $ln(x-6)+ln(x+3)=ln22$ ?

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How do you solve $ln (x - 6) + ln (x + 3) = ln 22$ $ln(x-6)+ln(x+3)=ln22$ ?

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Answer 1 · 2015-08-13T00:49:29

$x = 8$ $color(red)( x=8)$

Explanation:

$ln (x - 6) + ln (x + 3) = ln 22$ $ln(x-6)+ln(x+3)= ln22$

Recall that $ln a + ln b = ln a b$ $lna+lnb=lnab$ .

∴ $ln (x - 6) (x + 3) = ln 22$ $ln(x-6)(x+3)=ln22$

Convert the logarithmic equation to an exponential equation.

$e^{ln ((x - 6) (x + 3))} = e^{ln 22}$ $e^ln((x-6)(x+3)) = e^ln22$

Remember that $e^{ln x} = x$ $e^lnx =x$ , so

$(x - 6) (x + 3) = 22$ $(x-6)(x+3)=22$

$x^{2} - 3 x - 18 = 22$ $x^2-3x-18=22$

$x^{2} - 3 x - 40 = 0$ $x^2-3x-40=0$

$(x - 8) (x + 5) = 0$ $(x-8)(x+5)=0$

$x - 8 = 0$ $x-8=0$ and $x + 5 = 0$ $x+5=0$

$x = 8$ $x=8$ and $x = - 5$ $x=-5$ are possible solutions.

Check:

$ln (x - 6) + ln (x + 3) = ln 22$ $ln(x-6)+ln(x+3)= ln22$

If $x = 8$ $x=8$ ,

$ln (8 - 6) + ln (8 + 3) = ln 22$ $ln(8-6)+ln(8+3)=ln22$

$ln 2 + ln 11 = ln 22$ $ln2+ln11=ln22$

$ln (2 \times 11) = ln 22$ $ln(2×11)=ln22$

$ln 22 = ln 22$ $ln22=ln22$

$x = 8$ $x=8$ is a solution.

If $x = - 5$ $x=-5$ ,

$ln (- 5 - 6) + ln (- 5 + 3) = ln 22$ $ln(-5-6)+ln(-5+3) =ln22$

$ln (- 11) + ln (- 2) = ln 22$ $ln(-11)+ln(-2)=ln22$

But $ln (- 11)$ $ln(-11)$ and $ln (- 2)$ $ln(-2)$ are not defined.

∴ $x = - 5$ $x=-5$ is not a solution.

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How do you solve $ln (x - 6) + ln (x + 3) = ln 22$ $ln(x-6)+ln(x+3)=ln22$ ?

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Explanation:

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How do you solve ln(x−6)+ln(x+3)=ln22ln(x-6)+ln(x+3)=ln22?

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Explanation:

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How do you solve $ln (x - 6) + ln (x + 3) = ln 22$ $ln(x-6)+ln(x+3)=ln22$ ?