Prove that (a^x)-x>1, when a>0; a is different from 1.
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Let f(x)=(a^x) -x -1
f(x)>f(0). From this statement, we have x=0 as the minim point.
f(x) has derivative because f(x) consists of elementary functions and for that is a continuing function which admits derivative.
Because f(x) is allowing derivative, we can apply Fermat's theorem. We just found out that x=0 is a minim point and due to the Fermat's theorem, f'(0)=0 (we know that an extrem point is that one which cancel f'(x))
f'(x)=[(a^x) -x -1]'= (a^x*lna)-1
f'(0)=a^0*lna -1=lna -1
But f'(0)=0, lna -1=0, ln a=1, a=e^1, a=e
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