How do you solve $ln (x + 1) - ln x = 5$ $ln(x+1)-lnx=5$ ?

Question

7462 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-09T14:01:05

See below.

When we subtract two natural logs (or two logs in general), we divide the expressions on the inside.

$ln (x + 1) - ln x = 5$ $ln(x+1)-lnx=5$

$ln (\frac{x + 1}{x}) = 5$ $ln((x+1)/x)=5$

$e^{5} = \frac{x + 1}{x}$ $e^5=(x+1)/x$

$x e^{5} = x + 1$ $xe^5=x+1$

$x e^{5} - x = 1$ $xe^5-x=1$

$x (e^{5} - 1) = 1$ $x(e^5-1)=1$

$x = \frac{1}{e^{5} - 1}$ $x=1/(e^5-1)$

Answer 2 · 2017-06-09T14:02:37

I got: $x = \frac{1}{e^{5} - 1} = 0.006783$ $x=1/(e^5-1)=0.006783$

We can use a property of logs:

$log x - log y = log (\frac{x}{y})$ $logx-logy=log(x/y)$

and write:

$ln (\frac{x + 1}{x}) = 5$ $ln((x+1)/x)=5$

use the definition of (natural) log:

$\frac{x + 1}{x} = e^{5}$ $(x+1)/x=e^5$

rearrange:

$x + 1 = x e^{5}$ $x+1=xe^5$

$x (e^{5} - 1) = 1$ $x(e^5-1)=1$

$x = \frac{1}{e^{5} - 1} = 0.006783$ $x=1/(e^5-1)=0.006783$

How do you solve ln(x+1)-lnx=5ln(x+1)−lnx=5ln(x+1)-lnx=5?