Question about logs

Question

For what values of $k$ will $kx^2 = \log(x)$ have exactly $1$ solution?

Received this question from a friend who was asked this in an interview. I thought the answer was $k=0$ but this was wrong.

Answer 1

HINT

We can distiguish three cases

• for $k=0$ we have exactly one solution for $x=1$

• for $k<0$ we have alway one intersection, it can be proved by studing sign and monoticity of the function $f(x)=\log x -kx^2$

• for $k>0$ we have one intersection when $\log x$ and $kx^2$ are tangent

Indeed since for k=0 we have one intersection point and for $k=1$ we have that

• $x^2>x-1\ge \log x$

there exists an intermediate value for $k$ such that the two curves are tangent.

The condition of tangency at $(x_0,y_0)$ can be expressed by

• tangent for $\log x$: $y-\log x_0=\frac1{x_0}(x-x_0)$
• tangent for $kx^2$: $y-kx_0^2=2kx_0(x-x_0)$

and thus we must have

• $\frac1{x_0}=2kx_0\implies k=\frac1{2x_0^2}$
• $\log x_0=kx_0^2=\frac12\implies x_0=\sqrt e \quad k=\frac1{2e}$

Thus for $k>0$ $\log x=kx^2$ has exactly one solution for $k=\frac1{2e}$ with tangent point at $(\sqrt e,\frac12)$.

Plot for the tangency case k>0

Answer 2

There is one $k$ for which the two (real) curves touch. It is determined by the condition that at the point $x_0$ where $kx_0^2=\ln x_0$, the slopes $2kx_0$ and $\frac1{x_0}$ are the same - which means that we must have $\ln x_0=kx_0^2=\frac12$, $x_0=\sqrt e$ and ultimately $$k=\frac 1{2e}.$$ As said, the $k$ will leead to a single touching point between the curves. For any $k>\frac1{2e}$, there will be no intersection at all. For any $k$ with $0<k<\frac1{2e}$, the will be two intesections (because the paraola eventually wins over the log). But for $$k\le 0,$$ we will have exactly one intersection as well.