If I have ln(elne) do I solve this as

lne + lne = 1 +1 = 2

or

ln(lne^{e}) = ln(e) = 1

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- Apr 8th 2013, 12:56 PMQuixoticLog and e problem
If I have ln(elne) do I solve this as

lne + lne = 1 +1 = 2

or

ln(lne^{e}) = ln(e) = 1 - Apr 8th 2013, 01:03 PMemakarovRe: Log and e problem
The first variant should say ln(e) + ln(ln(e)) = 1 + ln(1) = 1 + 0 = 1. The second variant is correct.

- Apr 20th 2013, 12:03 AMBradynsRe: Log and e problem
You could just treat it like this.

Equation (1)

$\displaystyle ln(eq)$, where $\displaystyle q=ln(e)$

Given that,

$\displaystyle q=ln(e)\Rightarrow q= 1$

Equation (1) can be rewritten as:

$\displaystyle ln(eq) \Rightarrow ln(1e) \Rightarrow ln(e) = 1$