What is ln^e(-eln(e^2)) ?

The solution to ln^e(-eln(e^2)) is ln^e(2)+1

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simplify ln^e(-eln(e^2))

simplify

ln^{e} (- e ln (e^{2}))

ln^{e} (2) + 1

Decimal Notation

1.36924 \dots

Solution steps

ln^{e} (- e ln (e^{2}))

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln^{e} (- e ln (e^{2})) = ln^{e} (e) + ln^{e} (ln (e^{2}))

= ln^{e} (e) + ln^{e} (ln (e^{2}))

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (e^{2}) = 2 ln (e)

= ln^{e} (e) + ln^{e} (2 ln (e))

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln^{e} (2 ln (e)) = ln^{e} (2) + ln^{e} (ln (e))

= ln^{e} (e) + ln^{e} (2) + ln^{e} (ln (e))

ln^{e} (e) = 1

ln^{e} (ln (e)) = 0

= 1 + ln^{e} (2) + 0

Add the numbers:

1 + 0 = 1

= ln^{e} (2) + 1

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