Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to work out the value of $k$, where $k>0$, for which $e^x=kx$ has $1$ solution. I've done it somewhat intuitively as follows:


By inspection we can see that when $x=1$, the exponent in the LHS and the multiplier(?) in the RHS are irrelevant, leaving us with


Therefore $e^x=kx$ has $1$ solution at $x=1$.

This is not a very rigorous solution at all though. What is a more proper way to solve this?

share|cite|improve this question
up vote 2 down vote accepted

The idea is that the two functions $f(x)=e^x$ and $g(x) = kx$ touch only once; that is, $g$ is tangent to $f$ at some point.

Denote the tangent point $x_0$. Then $f(x_0)=e^{x_0}=kx_0=g(x_0)$. The slopes of the two functions have to be the same at this point as they are tangent: $$f'(x_0) = g'(x_0)$$ $$e^{x_0} = k$$ and so $$ kx_0 = k$$ The only solution is $x_0=1$ and the value of $k$ is then $k=e^{x_0} = e$.

Postscript: Negative $k$

I assumed above that $k$ is positive. For $k$ negative we can observe that $$ \lim_{x \rightarrow -\infty} (e^x - kx) = -\infty $$ and $$ \lim_{x \rightarrow +\infty} (e^x - kx) = +\infty $$ so that by the intermediate value theorem there exists a point at which $e^x-kx=0$. The function $h(x)=e^x-kx$ is monotonically increasing, as $h'(x)=e^x-k>0$, so there is only one point at which $e^x=kx$.

share|cite|improve this answer
And here's a plot:… :) – James Fennell Dec 4 '12 at 20:09
I think this lacks rigour. In particular, why does the existence of a unique solution imply that both kx is a tangent to $e^x$ – Amr Dec 4 '12 at 20:11
Thanks a lot, it didn't occur to me that you could get 2 simultaneous equations from the fact that the gradients are equal and the functions are equal :) – Taimur Dec 4 '12 at 20:16
@Amr, I think it would be sufficient to sketch the graphs and show that since both functions have positive gradients, one intersection guarantees a tangent. – Taimur Dec 4 '12 at 20:17
No it is not sufficent. $k=-1$ is a possible value – Amr Dec 4 '12 at 20:19

Since $e^x>0$ for all real $x$, then $y=e^x$ does not intersect $y=kx$ at all if $k=0$.

If $k<0$, then observe that $e^x-kx$ is strictly increasing, with $$\lim_{x\to\infty}(e^x-kx)=\infty$$ and $$\lim_{x\to-\infty}(e^x-kx)=-\infty.$$ From the strict monotonicity, there is at most one solution to $e^x=kx$, and from an application of the Intermediate Value Theorem, there is at least one solution.

Suppose that $k$ is positive and that $y=kx$ intersects $y=e^x$ at exactly one point--equivalently, that $f(x)=e^x-kx$ has exactly one zero. Now, $f'(x)=e^x-k$, and by observing the sign of $f'(x)$, we conclude that $f$ is decreasing on $(-\infty,\ln k)$ and increasing on $(\ln k,\infty)$, achieving a global minimum when $x=\ln k$. Noting that $f(x)\to\infty$ as $x\to\pm\infty$, it follows that the minimum value of $f(x)$ cannot be negative, for otherwise, $f(x)$ would have two zeroes--one in $(-\infty,\ln k)$ and one in $(\ln k,\infty)$--but on the other hand, the minimum value of $f(x)$ cannot be positive, either, for otherwise $f(x)$ would have no zeroes. Thus, the minimum value of $f(x)$ (which, recall, is achieved at $x=\ln k$) is $0$, meaning $$0=f(\ln k)=e^{\ln k}-k\ln k=k-k\ln k=k(1-\ln k).$$ Since $k>0$, this means that $1-\ln k=0$, so $\ln k=1$, and so $k=e$.

share|cite|improve this answer

You want the two curves to be tangent so the slope at the common point is the same. This requires a common solution to $e^x=kx$ and $e^x=k$. Equating the right sides, we get $kx=k$, so $x=1$ or $k=0$, but $k=0$ is not allowed, so $x=1, k=e$

share|cite|improve this answer
Why does $kx$ have to be a tangent to $e^x$. In fact, k=-1 is a possible value. – Amr Dec 4 '12 at 20:24
Even if k is positive, I don't see why the fact that the slopes are equal follows immediately from the fact that there is only one solution – Amr Dec 4 '12 at 20:33
@Amr: $e^x \gt kx$ at both $-\infty$ and $+\infty$, so if they are not tangent there will be two points of intersection. – Ross Millikan Dec 4 '12 at 20:37
Now I see.Comments must be at least 15 characters in length.(click on this box to dismiss) – Amr Dec 4 '12 at 20:39

It can be shown that for all $0<k<e$, $e^x=kx$ has no solution. (This is done using calculus, by showing that $e^x-kx$ has a minimum of $k-klog(k)$. Since $0<k<e$, we know that $log(k)<1$, hence $k-klog(k)>0$)

If $k>e$, we know that $e^x-kx$ will be negative for some $x$ (At $x=log(k)$). Since, $e^x-kx$ grows without bound as $x$ approaches infinity or negative infinity. We can show that $e^x-kx$ crosses the x axis twice (a contradiction)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.