Real Numbers (1 Viewer)

[locked.on]

Quick question:

For which positive real numbers a is it true that a^x ≥ 1+x for all real x?

Thanks in advance.

 

[Affinity]

e

 

[locked.on]

How so?

What's the working or proof behind the answer?

 

[lolokay]

I think it would be because the tangent of e^x at x=0 is 1, whereas this is not true for any other number (the tangent of a^x at x=0 is ln a]).

The y value at x=0 will be 1 for the exponential function. The tangent of x+1 is always 1, and y=1 for x=0. So, if dy/dx is lower than 1 at x=0, as x increases, y decreases at a lesser rate than x+1 and so 'sinks' beneath it. If dy/dx is greater than 1 at x=0, then as x decreases, y will decrease at a greater rate than in x+1, and again it will 'sink' beneath it.

So the tangent at x=0 must be 1, which is true only for e.