How do you solve $ln (x + 4) - ln (x + 3) = ln x$ $ln(x+4)-ln(x+3)=lnx$ ?

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Answer 1 · 2016-03-23T23:20:48

$x = - 1 + \sqrt{5}$ $x=-1+sqrt(5)$

Dominion: $x + 4 > 0 and x + 3 > 0 and x > 0 \Leftrightarrow x > 0$ $color(blue)(x+4>0 and x+3>0 and x>0 <=> x>0$

$ln (x + 4) - ln (x + 3) = ln (x)$ $ln(x+4)-ln(x+3)=ln(x)$

$ln (x + 4) = ln (x + 3) + ln (x)$ $ln(x+4)=ln(x+3)+ln(x)$

The sum of logarithms is the logarithm of the product

$ln (x + 4) = ln ((x + 3) x)$ $ln(x+4)=ln((x+3)x)$

$ln (a) = ln (b) \to a = b$ $color(blue)(ln(a)=ln(b) ->a=b$

$x + 4 = (x + 3) x$ $x+4=(x+3)x$

$x + 4 = x^{2} + 3 x$ $x+4=x^2+3x$

$x^{2} + 2 x - 4 = 0$ $x^2+2x-4 =0$

$x = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (- 4)}}{2}$ $x=(-2+-sqrt(2^2-4*1*(-4)))/2$

$x = \frac{- 2 \pm \sqrt{4 + 16}}{2}$ $x=(-2+-sqrt(4+16))/2$

$x = \frac{- 2 \pm \sqrt{20}}{2}$ $x=(-2+-sqrt(20))/2$

$x = \frac{- 2 \pm 2 \sqrt{5}}{2}$ $x=(-2+-2sqrt(5))/2$

$x = - 1 \pm \sqrt{5}$ $x=-1+-sqrt(5)$

$x = - 1 + \sqrt{5}$ $x=-1+sqrt(5)$ , because x must be larger than 0

How do you solve ln(x+4)−ln(x+3)=lnxln(x+4)-ln(x+3)=lnx?