ln e^1/2=1/2 How did you get it

ln e^1/2=1/2 How did you get it


Lne ^ 1 / 2 = 1 / 2 * lne = 1 / 2, where:
1) Ln (x ^ n) = n * LNX, the algorithm of logarithm
2) Lne = 1, LN is the logarithm with base E



How to solve y = ln (TaNx / 2) - ln1 / 2


Remember a formula [V (U)] '= u'v' (U), that is ln (TaNx / 2) + ln1 / 2 = (1 / (2cosx ^ 2)) * (2 / TaNx), where ln1 / 2 is a constant and the derivative is zero



-Is ln (x) = ln1 / x
Is - LNX equal to ln1 / X and why?


Equal to
Because - LNX = ln (x ^ - 1) and x ^ - 1 = 1 / x, - LNX = ln (1 / x)

Elden Ring Premiere

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