For which values ​​of a equation $y=ax^2$ is tangent to $y=\ln(2x)$ For which values ​​of a equation $y=ax^2$ is tangent to $y=\ln(2x)$
Hello, I tried to solve this question and I can not continue.
$$y=ax^2\\
y=\ln(2x)$$

What I did so far is  to $d/dx$ of $y=ax^2$ $\to$ $2ax=\ln(2x)$, now I dont know if to $d/dx(\ln(2x))$

What you are suggesting? 
Thanks.
 A: We want the tangent line to $y=ax^2$ at some point $(p,q)$ to be the same line as the tangent line to $y=\ln(2x)$ at $(p,q)$.
The tangent line to $y=ax^2$ at a general point $(x,y)$ on the curve has slope $\frac{d}{dx}(ax^2)=2ax$.  Similarly, the tangent line to $y=\ln(2x)$ has slope $\frac{2}{2x}=\frac{1}{x}$.
For the tangent lines to the two curves at $(p,q)$ to be the same,  we need the slopes to be the same, so we need
$$2ap=\frac{1}{p}.\tag{$1$}$$
The point $(p,q)$ must be on both curves, so we also need
$$q=ap^2\qquad\text{and}\qquad q=\ln(2p).\tag{$2$}$$ 
We have three equations in $3$ unknowns $a$, $p$, and $q$, and need to solve for $a$.
From Equation $(1)$ and the first equation of $(2)$, we get $q=\frac{1}{2}$. Then from the second equation in $(2)$ we get $\ln(2p)=\frac{1}{2}$ and therefore $p=\frac{1}{2}e^{p/2}$,
Finally, since from $(1)$ we have $a=\frac{1}{2p^2}$, we can calculate $a$.  
A: There are 4 solutions (4 intersecting points):


*

*$a \neq 0$  and $Im(W_{-1}(\frac{-a}{2})) \gt -2 \pi$ and $x=\frac{1}{2} e^{\frac{-1}{2}W_{-1}(\frac{-a}{2})}$ and $y=\frac{1}{4}a$ $e^{W_{-1}(\frac{-a}{2})}$

*$a \neq 0$ and $x=\frac{1}{2} e^-{\frac{W(\frac{-a}{2})}{2}}$

*$a\neq0$ and $Im(W_{-1}(\frac{-a}{2})) \lt 2\pi$ and $x=\frac{1}{2}e^{\frac{-1}{2}W_1(\frac{-a}{2})}$ and $y=\frac{1}{4}ae^{-W_1(\frac{-a}{2})}$

*$a=0$  and $x=\frac{1}{2}$ and $y=0$
where $W_k$ is Lambert W Function.
A: I'm going to throw my hat in here, because there are really only two unknowns and two equations.  The curves are tangent at their one intersection point in the first quadrant.  We want the functions to be equal there:  
$$a x^2 = \ln(2x),$$
and their slopes to be equal there as well:
$$2 a x  =  \frac{1}{x}.$$
The derivatives equation gives us $x^2 = \frac{1}{2a}$ .  Inserting that into the functions equation yields
$$ a \cdot (\frac{1}{2a}) = \ln( 2 \cdot (\frac{1}{2a})^{1/2}) \Rightarrow  \frac{1}{2} = \ln 2 - \frac{1}{2}\ln(2a) $$
$$\Rightarrow \ln a = (2 \ln 2) - (\ln 2) - 1  \Rightarrow  a = e^{(\ln 2) - 1 }  =  \frac{2}{e}  \approx 0.7358 .   $$
A graph in fact appears to confirm this. 
The tangent point (x,y) does not have a tidy value; I haven't worked that out yet, as it was not asked for.
EDIT: The defining equation for  $x$  is  $x^2 - \frac{e}{2} \ln x  =  \frac{e}{2} \ln 2$ .  [A little time with Newton's method indicates that the tangent point is approximately $(0.805,0.477)$ .]
ADDENDUM (made a few hours later):  I got curious about a generalization of this.  For $y = ax^n$ tangent to $y = \ln (kx)$  [$n$ and $k$ being positive integers], we find that $a = \frac{k^{n}}{ne}  $, and the location of the tangent point is found from 
$$x^n - (\frac{ne}{k^n}) \cdot \ln x = \frac{ne}{k^n} \cdot \ln k  .  $$  
