0

I am trying to prove that $y=2^x$ and its inverse $y=log_{2}{x}$ do not intersect.

If $y=a^x$, then $\frac{dy}{dx}=a^xlna$

Solving $a^xlna=1$ we get $x=\frac{1}{lna}ln(\frac{1}{lna})$.

So for $y=x$ to be a tangent to $y=a^x$, $\frac{1}{lna}ln(\frac{1}{lna})=\frac{1}{lna}$ which then gives $a=e^{1/e}$.

Now since $y=x$ is a tangent to $y=(e^{1/e})^x$ at $(e,e)$ and $2^x>(e^{1/e})^x\geq x$ for all positive $x$, we conclude that $y=2^x$ and $y=x$ do not intersect and hence $y=a^x$ does not intersect its inverse as they are supposed to be symmetrical about $y=x$

For negative $x$ or when $x=0$ it is trivial.

Is there a simpler way?

LanaDR
  • 503

2 Answers2

3

Using $2^x > x$ for all $x \in \mathbb{R}$ you get taking the logarithm $$2^x >x \Rightarrow x > \log_2 x \stackrel{x>0}{\Rightarrow} 2^x > x > \log_2 x$$

trancelocation
  • 32,243
0

For sure, not simpler !

Suppose that you consider the function $$f(x)=2^x-\frac{\log (x)}{\log (2)}$$ Computing the derivatives $$f'(x)=2^x \log (2)-\frac{1}{x \log (2)}$$ $$f''(x)=\frac{1}{x^2 \log (2)}+2^x \log ^2(2) \qquad >0 \qquad \forall x >0$$ The first derivative cancels at $$x_*=\frac{W\left(\frac{1}{\log (2)}\right)}{\log (2)}$$ where appears Lambert function that you can evaluate using the series expansion given in the linked page to get $x_*\approx 1.02372$ and this is a minimum by the second derivative test.

Since $x=1+\epsilon$, to be accurate, let us expand as a Taylor series to get locally $$f(x)=2+(x-1) \left(2 \log (2)-\frac{1}{\log (2)}\right)+O\left((x-1)^2\right)$$ giving $f(x_*)\approx 1.999$.

Claude Leibovici
  • 260,315