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

Question

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...

Answer 1

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 $1 <b <b_0$. 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 $b=b_0$

[

Answer 2

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 $b = e^{1/e} \sim 1.444667861$.