Intersection of two graphs

If you read the other solutions, you will notice that a lot of people made the assumption that:

Because $f(x) = x^2$ and $g(x) = \log_b x$ has exactly one solution at $x = x^*$ , hence they must share a common tangent at $x^*$ .

Unfortunately, this claim need not be true. Lots of curves can intersect once, and not share a common tangent at the point of intersection. Can you find various counterexamples (of essentially different scenarios) where there isn't a common tangent?

Hint: Think about your favourite function $h(x)$ where $h(x) = 0$ has a unique solution at $x = x^*$ and $h'(x^*) \neq 0$ . Then, set $f(x) = h(x) + g(x)$ .
Note: $h(x)$ is ultimately the crux to analyze.

---

So, when must we have a common tangent? What additional assumption about these functions is being used? Can you list some in the comments?

More concretely, what are the conditions that when placed on a (differentiable) function $h(x) : [0,1] \rightarrow \mathbb{R}$ that will allow us to conclude that $h'(x^*) = 0$ ? By setting $h(x) = f(x) - g(x)$ , we reach our goal of determinining when the uniqueness of a solution implies a common tangent $f(x) = g(x), f'(x) = g'(x)$ .

Example 1 : If $h(x) = 0$ has a unique solution at $x=x^*$ and $h(0) h(1) > 0$ , then $h'(x^*) = 0$ .

Proof: $h(0) \times h(1) > 0$ means that they have the same sign. WLOG, suppose that $h(0), h(1) > 0$ .
Proof by contradiction. Suppose $h'(x^*) \neq 0$ . WLOG, suppose that $h' (x^*) < 0$ .
Then, there exists an $\epsilon > 0$ such that $\frac{ h(x^* + \epsilon) - h(x^*) } { \epsilon } < 0 \Rightarrow h(x^* + \epsilon) < 0$ .
Then, by the intermediate value theorem, we can conclude that in the interval $[x^*+ \epsilon, 1 ]$ , there must be another solution to $h(x) = 0$ .
This contradicts that there is a unique solution in $[0,1]$ .

Note: The condition of $[ f(0) - g(0) ] \times [f(1) - g(1) ] > 0$ is typically the one that people think of in such a scenario, but then have difficulty expressing it clearly.
Note: The condition doesn't need to apply to the endpoints of the function's domain. It just needs to apply to an interval containing the unique solution. Note: In case it wasn't obvious, the conditions are sufficient, but not necessary. IE Just because the function fails the conditions doesn't mean that we must have $h'(x^*) \neq 0$ .

Corollary: This applies to the given problem.

---

(Suggested by Henry) Example 2 : If $h(x) = 0$ has a unique solution at $x=x^*$ and $h'(x) > 0$ on $( x^*, 1 ],$ , then $h'(x^*) = 0$ .

This doesn't work. Can you come up with a counterexample? Can you tweak this condition so that it will work?

(Suggested by Antoine) Example 3 : If $h(x) = 0$ has a unique solution at $x=x^*$ and $h''(x) > 0$ , then $h'(x^* ) = 0$ .

This doesn't work. Can you come up with a counterexample? Can you tweak this condition so that it will work?

The equation $x^2 = \log_bx$ can be rewritten as $b = x^{\frac{1}{x^2}}$ . If we consider the function $y \; = \; x^{\frac{1}{x^2}} \hspace{2cm} y' \; = \; x^{\frac{1}{x^2}-3}(1 - 2\ln x)$ we see that $y$ takes its maximum at $x = \sqrt{e}$ , and the maximum value of $y$ is $b = e^{\frac{1}{2e}}$ . This makes the answer $\boxed{2019}$ .

The question is equivalent to ask that $h(x) = f(x)-g(x)$ has a single zero. Note that $h$ is a continuous differentiable function on the interval $[0,+\infty[$ (since $x^2$ and $\log_b$ are). Furthermore $h$ tends to $\infty$ as $x \to 0^+$ or $x \to +\infty$ . For a such a function to have a single zero means that $h(x) =0$ and $h'(x) =0$ (see Lemma below).

$h(x) =0$ is the equation (I): $x^2 - \frac{\ln x}{\ln b} =0$ and $h'(x) =0$ is the equation (II): $2x - \frac{1}{x \ln b} =0$ .

(II) $\iff 2x = \frac{1}{x \ln b} \iff 2x^2 = \frac{1}{\ln b} \iff x^2 = \frac{1}{2\ln b}$ .

You can input that in (I): $x^2 - \frac{\ln x}{\ln b} =0 \iff \frac{1}{2\ln b} - \frac{\ln x}{\ln b} =0 \iff \frac{1}{2} = \ln x \iff x = \sqrt{e}$ .

Now use (II) again: $e = \frac{1}{2\ln b} \iff b = e^{1/2e}$ .

Lemma (basically just the intermediate value theorem)

Statement of the Lemma: Assume $h$ has domain $[a,c]$ (the interval could be half-open or open; $a$ and $c$ could be infinite), is such that $\displaystyle \lim_{x \to a} h(x)= >0$ (with $+ \infty$ being considered $>0$ ), $\displaystyle \lim_{x \to c} h(x) >0$ and $h$ has a derivative everywhere. Then if $h$ has only one zero on $[a,c]$ , the derivative of $h$ also has a zero there.

Proof of the Lemma: Assume the function has only one zero $x_0$ and that $h'(x_0) \neq 0$ . Then there are $x$ very close to $x_0$ so that $h(x_0) <0$ . Let $x_1$ be such a point, then by the intermediate value theorem there are two zeroes: one in the interval $[a,x_1]$ and one in the interval $[x_1,c]$ .

$x^2=\frac {\ln x}{\ln b}$

(Let $\ln b=c$ )

The graphs of these two functions will have exactly one solution when they just touch. Thus, not only will the function value be equal but also the value of slope of the tangent at that point (let it have abscissa $x_1$ )

$\therefore, f(x_1)=g(x_1);f'(x_1)=g'(x_1)$

$2x_1=\frac {1}{x_1c}$

$x_1=\frac {1}{\sqrt {2c}}$

Putting that in the very first equation, we get $c=\frac {1}{2e}$

Hence, $b=e^{\frac {1}{2e}}≈1.2019$

Thus, $\lfloor10000 \{ b\} \rfloor =2019$

We know that $x-1$ is tangent to $\ln(x)$ . This is close to what we want, so let's work with it. Let's add $1$ , getting that $x$ is tangent to $\ln(ex)$ . That's closer - let's substitute $x^2$ for $x$ , giving us that $x^2$ is tangent to $\ln(ex^2)$ . Multiplying by $e$ and substituting $x/\sqrt e$ for $x$ gets us that $x^2$ is tangent to $e\ln(x^2)$ , and simplifying gives us that $x^2$ is tangent to $2e\ln(x)$ .

Finally, log identities give us $2e\ln(x)=\log_{e^{1/2e}}(x)$ .

Let's consider the difference function $y = f(x) = x^2 - log_b(x) = x^2 - \frac{ln(x)}{ln(b)}$

We'll be studying this function on the domain: $(0, \infty]$ .

Let's study it derivative:

$\frac{\mathrm{d} y}{\mathrm{d} x} = 2x - \frac{1}{x \cdot ln(b)}$

Let's find the extrema of f.

$\frac{\mathrm{d} y}{\mathrm{d} x} = 2x - \frac{1}{x \cdot ln(b)} =0$

$2x = \frac{1}{x \cdot ln(b)}$

$x = \frac{1}{ \sqrt {2 ln(b)}}$

Let's re-inject this to f.

$y_m = \frac{1}{ 2 ln(b)} - \frac{ln( \frac{1}{ \sqrt {2 ln(b)}})} {ln(b)}$

$y_m = \frac{ 1 + ln(2) + ln(ln(b)) }{2 ln(b)}$

As $x \rightarrow 0$ , $y \rightarrow \infty$ . As $x \rightarrow \infty$ , $y \rightarrow \infty$ .

So we can see that if $y_m$ is positive, then y=0 has no solutions. If $y_m$ is negative, y=0 has two solutions.

y=0 has a single solution if $y_m = 0$ .

$y_m = 0 \Leftrightarrow 1 + ln(2) + ln(ln(b)) = 0 \Leftrightarrow ln(b) = e^{-1 - ln(2)} = \frac{1}{2e} \Leftrightarrow b = e^{1/2e}$

$b = e^{1/2e} = 1.2019433684...$

$\lfloor 10000 b \rfloor = \boxed{2019}$

$x^2 = \dfrac {ln(x)}{ln(b)}$

Consider the two curves, $y = x^2$ and $y = \dfrac {ln(x)}{ln(b)} = kln(x)$

These two curves have exactly one solution implies that they have a common tangent i.e, both have same slope.

$\dfrac {dy}{dx} = 2x = \dfrac kx \Rightarrow k = 2x^2$

$x^2 = 2x^2 ln(x) \Rightarrow ln(x) = \dfrac 12 \Rightarrow x = e^{1/2}$

Now, $(e^{1/2})^2 = k ln(e^{1/2}) \Rightarrow k = \dfrac {e}{(\dfrac 12)} = 2e$

Now, $k = \dfrac {1}{ln(b)} \Rightarrow ln(b) = \dfrac 1k = \dfrac {1}{2e} \Rightarrow b = (e)^{1/2e} =1.20194336847$

$\lfloor 10000 \{ b \} \rfloor = \lfloor 10000\times 0.20194336847 \rfloor= \lfloor 2019.4336847\rfloor = \boxed {2019}$

Henry ,maybe you should have posted this question on New Year Eve ,Year 2019!!

A solution has $x^2=\log_b{x}=(\log_b{e})(\ln{x})=\frac{\ln{x}}{\ln{b}}$ so that $\ln{b}=\frac{\ln{x}}{x^2}$ and $b=e^{\frac{\ln{x}}{x^2}}$ .

Also, at the intersection point the derivatives have to be equal, because otherwise there will be two intersection points and solutions(*).

So $2x=\frac{\log_b{e}}{x} \Rightarrow 2x^2=\log_b{e} \Rightarrow \ln{b}=\frac{1}{2x^2}$ . Above we already saw that $\ln{b}=\frac{\ln{x}}{x^2}$ so that we can solve x from $\frac{\ln{x}}{x^2}=\frac{1}{2x^2} \Rightarrow \ln{x}= \frac{1}{2}  \Rightarrow x= \sqrt{e}$ .

Now $b=e^{\frac{\ln\sqrt{e}}{e}}=e^{{\frac{1}{2e}}} = 1.2019433..$ so that the requested answer is $\boxed{2019}$

A little bit early maybe, but Happy New year!

(*)This is because the second derivative of $x^2$ is 2 (positive), while the second derivative of $\log_b(x)$ is $-\frac{\log_b{e}}{x^2}$ (negative) on the entire (positive real) domain, as can also be seen from the graph. As $x\rightarrow \infty$ the tangent of the red curve goes to $+\infty$ and that of the blue curve goes to 0; whereas when $x \rightarrow 0^+$ , the tangent of the red curve goes to 0 and that of the blue curve goes to $-\infty$ .

Used WolframAlpha to obtain almost a real solution by adjusting value of 'b':

solve x^2 = log(x)/log(1.2019435)

x = 1.64872 + 0.000899216i

x = 1.64872 - 0.000899216i

The b=1.2019435 as above was bracketed as follows:

The two curves are either (1) cutting at two places giving two real roots or (2) not touching each other giving complex roots.

Therefore, imaginary part vanishing shows that two curves are touching each other at x = 1.64872.

The following answer of complex roots was obtained at b=1.20199 :

x = 1.64852 - 0.0169286 i

x = 1.64852 + 0.0169286 i

The following answer of two real roots was obtained at b=1.2019 :

x = 1.63258

x = 1.66524

Answer=2019

Try similar problems Common Point and Common Tangent .

$x^2=\log_{b} {x}=\dfrac{\ln{x}}{\ln{b}}$ , 2x=\dfrac{1}{x\ln{b} .

$x^2\ln{b}=\ln{x}=\dfrac{1}{2}$ , $x=\sqrt{e}$ .

Thus, $b=e^{\frac{1}{2e}}$ .