When x > 0, x > ln (1 + x)

When x > 0, x > ln (1 + x)

Analysis, to prove that x > ln (1 + x)
It is proved that x-ln (1 + x) > 0
prove:
Let t (x) = x-ln (1 + x)
Derivative t '= 1-1 / (1 + x)
=x/(1+x)>0
So, t is an increasing function,
t(x)>t(0)=0
Therefore, x-ln (x + 1) > 0
Therefore, x > ln (x + 1)

Elden Ring Premiere

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