How do you solve $ln x + ln (x + 1) = 2$ $lnx+ln(x+1)=2$ ?

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How do you solve $ln x + ln (x + 1) = 2$ $lnx+ln(x+1)=2$ ?

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Answer 1 · 2015-11-24T05:40:07

$x = \frac{- 1 + \sqrt{1 + e^{2}}}{2}$ $x = (-1 + sqrt(1+e^2))/2$

Explanation:

Before beginning, note that we do not accept values less than $0$ $0$ for the $ln$ $ln$ function, and thus as we are beginning with $ln (x)$ $ln(x)$ in the equation, it must be that $x > 0$ $x > 0$ .

We will use the following properties of logarithms:
(1) $ln (a b) = ln (a) + ln (b)$ $ln(ab) = ln(a) + ln(b)$
(2) $e^{ln (a)} = a$ $e^(ln(a)) = a$

as well as the quadratic formula:
$a x^{2} + b x + c = 0 \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $ax^2 +bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)$

By (1)
$ln (x) + ln (x + 1) = ln (x (x + 1)) = ln (x^{2} + x)$ $ln(x) + ln(x+1) = ln(x(x+1))=ln(x^2 + x)$
$\Rightarrow ln (x^{2} + x) = 2$ $=> ln(x^2 + x) = 2$

Then, by (2)
$e^{2} = e^{ln (x^{2} + x)} = x^{2} + x$ $e^2 = e^(ln(x^2+x)) = x^2 + x$
$\Rightarrow x^{2} + x - e^{2} = 0$ $=> x^2 + x - e^2 = 0$

Now we apply the quadratic formula with $a = b = 1$ $a=b=1$ and $c = - e^{2}$ $c=-e^2$

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

But as noted at the beginning, we must have $x > 0$ $x > 0$ and so we reject $\frac{- 1 - \sqrt{1 + e^{2}}}{2}$ $(-1-sqrt(1+e^2))/2$

Thus we are left with the answer

$x = \frac{- 1 + \sqrt{1 + e^{2}}}{2}$ $x = (-1 + sqrt(1+e^2))/2$

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