How do I solve logarithms?

Question

ln lxl =1

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Answer 1 · 2018-04-20T12:30:06

Please see below.

Here,

$ln | x | = 1 \Rightarrow {log}_{e} | x | = 1$ $ln |x|=1=>log_e|x|=1$

We know that,

${log}_{e} x = y \Leftrightarrow x = e^{y}, w h e r e, x > 0$ $log_ex=y<=>x=e^y,where, x > 0$

$:.log_e|x|=1=>|x|=e^1$

$i.e. |x|=e=>x=+-e$

Note:

$"ln (x) is undefined when"$ $x<=0"$

So, $ln|x|=ln|+-e|=ln e=1$

Answer 2 · 2018-04-20T12:30:28

$x=+-e$

We use the definition:

$log_b x = y iff b^y = x$

So for the given expression:

$ln|x|=1 iff e^1 = |x|$

$:. |x| = e => x =+-e$