$\ln\left(e^a\right)

a

=

\ln\left(e^a\right)

b e r \overset{a}{¨} k n a r d e n [[E x p o n e n t * W o r d l i s t * ∣ e x p o n e n t]] m a n s k a s \overset{a}{¨} t t a p \overset{a}{˚} b a s e n

e

f \overset{o}{¨} r a t t p o t e n s e n s v \overset{a}{¨} r d e s k a b l i

e^a

. D e n e x p o n e n t e n k a n a v l \overset{a}{¨} s a s d i r e k t f r \overset{a}{˚} n u t t r y c k e t s o m

a.

S o m e x e m p e l b l i r d e n n a t u r l i g a l o g a r i t m e n a v

e^3

l i k a m e d

3,

e

u p p h \overset{o}{¨} j t t i l l " o c h "

\ln$" tar ut varandra]].