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, 01:56 PM #1

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- Apr 8th 2013, 02:03 PM #2

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- Apr 20th 2013, 01:03 AM #3
## Re: 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:

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