How do you solve $ln x - ln (x + 1) = 1$ $Ln x - Ln(x+1) = 1$ ?

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How do you solve $ln x - ln (x + 1) = 1$ $Ln x - Ln(x+1) = 1$ ?

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Answer 1 · 2016-12-01T22:27:53

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Explanation:

Use the subtraction rule of logarithms that ${log}_{a} (n) - {log}_{a} (m) = {log}_{a} (\frac{n}{m})$ $log_a(n) - log_a(m) = log_a(n/m)$ .

$ln (\frac{x}{x + 1}) = 1$ $ln(x/(x + 1)) = 1$

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

$x = e (x + 1)$ $x = e(x + 1)$

$x = e x + e$ $x = ex + e$

$x - e x = e$ $x- ex = e$

$x (1 - e) = e$ $x(1 - e) = e$

$x = \frac{e}{1 - e}$ $x= e/(1 - e)$

If you want an approximation, $x ≅ - 1.58$ $x ~= -1.58$ . However, since $x > 0$ $x > 0$ for the natural logarithms to be defined, this solution is extraneous.

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How do you solve $ln x - ln (x + 1) = 1$ $Ln x - Ln(x+1) = 1$ ?

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