# use the intermediate value theorem to find all cin the interval [-5,5] such that `log5(x^2+16)=2` (log base 5... didn't see a log function available)

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You need to use the intermediate value theorem for the continuous function `log_5 (x^2 + 16) = 2` , over the interval `[-5,5]` , such that:

`(f(5) - f(-5))/(5 - (-5)) = f'(c)`

You should notice that the function `log_5 (x^2 + 16) = 2` is constant for all `x in [-5,5]` , hence `f(5) = f(-5) = 0` such that:

`(0 - 0)/(5+5) = f'(c) => f'(c) = 0`

You should differentiate the function `f(x) = log_5 (x^2 + 16) - 2 ` with respect to x, such that:

`f'(x) =(2x)/(ln 5*(x^2 + 16)) `

Evaluating `f'(c)` yields:

`f'(c) = (2c)/(ln 5*(c^2 + 16)) `

Solving for c the equation `f'(c) = 0` yields:

`(2c)/(ln 5*(c^2 + 16))= 0 => 2c = 0 => c = 0 in [-5,5]`

**Hence, evaluating `c in [-5,5]` , using the mean value theorem, for the given function `log_5 (x^2 + 16) = 2` , yields **`c = 0.`