# Given the function f(x) = 7^x+a^x-3^x-5^x, what is a >0 if f(x)>=0? (solve with Fermat's theorem)

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### 1 Answer

You should notice that the given function is differentiable over real set R and you also should notice that evaluating the function at `x = ` 0 yields:

`f(0) = 7^0 + a^0 - 3^0 - 5^0 => f(0) = 1 + 1 - 1 - 1 = 0`

The problem provides the information that `f(x)>=0` , hence, replacing f(0) for 0 yields:

`f(x)>=f(0) =>` the point (0,0) represents a relative minimum point of the given function

By Fermat's theorem yields that evaluating the value of derivative of the function at the critical value `x = 0, f'(0) = 0.`

You need to evaluate the derivative of the function, such that:

`f'(x) = 7^x*ln 7 + a^x*ln a - 3^x*ln 3 - 5^x*ln 5`

Replacing 0 for x in equation of derivative, yields:

`f'(0) = 7^0*ln 7 + a^0*ln a - 3^0*ln 3 - 5^0*ln 5`

`f'(0) = ln 7 + ln a - ln 3 - ln 5`

Since `f'(0) = 0` yields:

`ln 7 + ln a - ln 3 - ln 5 = 0 => ln 7 + ln a = ln 3 + ln 5`

Converting the summations of logarithms in logarithms of the products yields:

`ln(7*a) = ln(3*5)`

Equating the arguments ofÂ logarithms yields:

`7a = 15 => a = 15/7`

Checking the validity of the value `a = 15/7` yields that this value is accepted, since the problem provides the information that `a > 0` .

**Hence, evaluating the value of a, using Fermat's theorem, yields **`a = 15/7.`

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