# what is a>0 if f(x)>=1 where f(x)=e^x-ax?

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The problem provides the information that `f(x) >= 1` , hence, the function `f(x)` reaches its minimum value 1 at `x = c` , such that:

`f'(c) = 0`

You need to evaluate `f'(x)` , such that:

`f'(x) = e^x - a`

You need to solve for x the equation `f'(x) = 0` , such that:

`e^x - a = 0 => e^x = a`

Taking common logarithms both sides, yields:

`ln e^x = ln a => x*ln e = ln a => x = ln a`

Hence `x = c = ln a` represents the critical value of the function, such that `f(ln a) = 1` , such that:

`f(ln a) = e^(ln a) - a*ln a => f(ln a) = a - aln a`

`f(ln a) = 1 => a - aln a = 1`

Considering `a = 1` yields:

`1 - 1*ln 1 = 1 - 0 = 1 valid => a = 1 > 0`

**Hence, evaluating the value of a, under the given conditions, yields `a = 1` .**