# Solve for values of b>1, such that $b^x$ and $\log_b x$ intersect only once.

Thinking about this problem, you would have the f(x)= $$b^x$$ and its inverse touching each other once before diverging away from each other. So I'm thinking this means $$x=y=$$ $$b^x$$= $$\log_b x$$ . Not sure where to proceed from here...

• Once you know the answer is $\sqrt[e]{e}$ it's fairly straightforward to show it works - but I'm afraid I found that by trial and error and I'm not sure how to directly show that's the value you want. Dec 3, 2019 at 23:46

Adding to @Dr Zafar Ahmed's answer, A simpler approach is, if $$b>1$$ and $$b^x = \log_b(x)$$, we automatically have $$b^x = \log_b(x) = x$$. From the first and the last bit, we get $$b = x^{1/x}$$ (It is crucial here that b>1. See if you can think why).
The Graph of x^(1/x) looks like this. A bit of basic calculus tells us that the function reaches its maximum at $$e^{1/e}$$. But notice also, that the function is one-one at that exact point only. So, given any "$$y$$" value, the function always takes 2 $$x$$ values. So, this directly means that in our case, we obtain a unique $$b$$ only when $$b = e^{1/e} \sim 1.444667861$$.
Number of real roots of $$b^x=\log_b x, b>1, x>0~~~~(1)$$ $$f(x)=b^x-\log_b x \implies f(0)= +\infty,~ f(\infty)= +\infty.$$ So $$f(x)=0$$ will have no real roots or even number of real roots or one root (two coincident roots) critically if $$b=b_0$$ such that $$f(x_0)=0$$, when $${b_0}^x=\frac{1}{x_0}$$ (when $$y=b^x$$ touches $$y=\log_b x$$).
Next $$f'(x)=b^{x} \ln b- \frac{1}{x} \ln b>0 ~\text{for}~ b>1, x>0.$$ This means $$f'(x)$$ will have atmost one real root and this inturn will mean that $$f(x)=0$$ will have atmost two real roots. For exampole for $$b=\sqrt{2}$$, $$b^x=\log_b x$$ has two roots $$x=2,4$$. Eq.(1) can have exactly one root only when $$y=b^x$$ touches $$y=\log_b x$$ at $$x=x_0$$, when $$b=b_0.$$ such that $${b_0}^x=\frac{1}{x_0}$$.
Finlly Eq. (1) will have one real root if $$b=b_0$$, no real root if $$b>b_0$$ and two reak roots if $$1 . the value of $$b_0$$ turns out to be a little more than $$\sqrt{2}$$, namely $$b_0=1.444667861009766.$$ See the Fig. below for the critical stuation when the curves touch at $$x=e$$. $$y=b^x$$ is blue, $$y=\log_b x$$ is read and $$b=b_0$$