How do you solve $ln (x) + ln (x - 2) = ln (x + 4)$ $ln(x) + ln(x-2) = ln(x+4)$ ?

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Answer 1 · 2015-10-21T11:32:56

Use properties of logs to derive a quadratic equation in $x$ $x$ , one of whose roots is a valid solution to the original equation.

$x = 4$ $x = 4$

$ln (x + 4) = ln (x) + ln (x - 2) = ln (x (x - 2))$ $ln(x+4) = ln(x) + ln(x-2) = ln(x(x-2))$

So

$x + 4 = x (x - 2) = x^{2} - 2 x$ $x+4 = x(x-2) = x^2 - 2x$

Subtract $x + 4$ $x+4$ from both ends to get:

$0 = x^{2} - 3 x - 4 = (x - 4) (x + 1)$ $0 = x^2-3x-4 = (x-4)(x+1)$

So $x = - 1$ $x = -1$ or $x = 4$ $x = 4$

Discard $x = - 1$ $x = -1$ since both $ln (- 1)$ $ln(-1)$ and $ln (- 1 - 2)$ $ln(-1-2)$ are not defined.

So the only solution is $x = 4$ $x = 4$

How do you solve ln(x)+ln(x−2)=ln(x+4)ln(x) + ln(x-2) = ln(x+4)?