Slope Tangent Similarity at how many points do the tangents to the functions $y=3^x$ and $y=x^3$ have the same slope?
No idea how to approach this.
 A: We set the derivatives of the two functions equal to one another:
$$\frac{d}{dx}\;3^x=\frac{d}{dx}x^3$$
$$3^x\ln(3)=3x^2$$
By plotting, we see that the graphs of the derivatives intersect (at least) thrice, so they have at least 3 points where their tangents have the same slope (the points are: $x\approx-0.47$, $x\approx1.12$ and $x\approx2.77$).
A: The equation which makes the slopes equal is $$f(x)=3^x\log(3)-3x^2=0$$ This kind of equation cannot be solved in terms of elementary functions but solutions exist using Lambert function. In this particular case, the solutions are given by $$x_1=-\frac{2 W\left(\frac{\log ^{\frac{3}{2}}(3)}{2 \sqrt{3}}\right)}{\log (3)} \simeq -0.467974$$ $$x_2=-\frac{2 W\left(-\frac{\log ^{\frac{3}{2}}(3)}{2 \sqrt{3}}\right)}{\log (3)} \simeq 1.11887$$ $$x_3=-\frac{2 W_{-1}\left(-\frac{\log ^{\frac{3}{2}}(3)}{2 \sqrt{3}}\right)}{\log (3)}\simeq 2.76762$$.
Even if this looks complex to you, let me underline that Lambert function is extremely useful. In particular, the solutions of any equation of the form $$a + b x+c \log(d x +e)=0$$ can be expressed using Lambert function (jus for memory, Lambert and Euler worked together this function which recently gained major interest in many practical areas of science and technology).
A: Not to put too fine a point on it or be the bearer of evil tidings, but I believe tha Rainier van Es's answer is in error, and that there is at least one solution $x_1 < 0$, as argued below.  For $x > 0$, I'm not sure how many solutions there are, but I suspect it is $0, 1$ or $2$.  I don't have a complete answer quite yet, but consider:
$\dfrac{d(x^3)}{dx} = 3x^2, \; \text{not} \; 2x,\tag{0}$
and this leads to Rainier van Es's error.  Meanwhile, read this:
Let
$y_1(x) = 3^x = e^{(\ln 3)x} \tag{1}$
and
$y_2(x) = x^3; \tag{2}$
then
$y_1'(x) = \ln 3 e^{(\ln 3)x} \tag{3}$
and
$y_2'(x) = 3x^2. \tag{4}$
The graphs of $y_1'(x)$ and $y_2'(x)$ intersect at exactly one point $x_1 < 0$.  To see this, note that $y_2'(x)$ is a parabola, concave upward, with least value $0$, whereas $y_1'(x)$ is a monotonically increasing exponential curve with $y_1'(x) \to 0$ as $x \to -\infty$ and $y_1'(x) \to \infty$ as $x \to \infty$ and $y_1'(0) = \ln 3$; it is easy to see, by envisioning or drawing these graphs that there must be $x_1$ as stated.  Or if a more rigorous approach is needed, consider for example $f(x) = y_2'(x) - y_1'(x)$ on the set $\infty < x \le 0$; for sufficiently large negative $x_0$, $f(x_0) > 0$ while $f(0) = -\ln 3$; by the intermediate value theorem we must have $x_1$, $x_0 < x_1 < 0$ with $f(x_1) = 0$; furthermore $f'(x) = 6x - (\ln 3)^2 e^{(\ln 3)x} < 0$ on $(-\infty, 0]$ so $f(x)$ can have at most one zero $x_1$ for $x \le 0$.
For $x \in [0, \infty)$, the situation is more complicated and I don't have a complete solution at this point.  Maybe someone can chime in with the details of the case $x > 0$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
