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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