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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:

$e^x=kx$

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

$e=k$

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?

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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 $e^x=kx$.

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  • $\begingroup$ And here's a plot: wolframalpha.com/input/… :) $\endgroup$ Dec 4 '12 at 20:09
  • $\begingroup$ I think this lacks rigour. In particular, why does the existence of a unique solution imply that both kx is a tangent to $e^x$ $\endgroup$
    – Amr
    Dec 4 '12 at 20:11
  • $\begingroup$ 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 :) $\endgroup$
    – Taimur
    Dec 4 '12 at 20:16
  • $\begingroup$ @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. $\endgroup$
    – Taimur
    Dec 4 '12 at 20:17
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    $\begingroup$ No it is not sufficent. $k=-1$ is a possible value $\endgroup$
    – Amr
    Dec 4 '12 at 20:19
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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 $k=e$.

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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 $x=1, k=e$

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  • $\begingroup$ Why does $kx$ have to be a tangent to $e^x$. In fact, k=-1 is a possible value. $\endgroup$
    – Amr
    Dec 4 '12 at 20:24
  • $\begingroup$ 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 $\endgroup$
    – Amr
    Dec 4 '12 at 20:33
  • $\begingroup$ @Amr: $e^x \gt kx$ at both $-\infty$ and $+\infty$, so if they are not tangent there will be two points of intersection. $\endgroup$ Dec 4 '12 at 20:37
  • $\begingroup$ Now I see.Comments must be at least 15 characters in length.(click on this box to dismiss) $\endgroup$
    – Amr
    Dec 4 '12 at 20:39
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An alternative to the methods presented so far is to explicitly solve the equation for $x$ in terms of the Lambert W function, $\text{W} (x)$.

Rewriting the equation as $$-x e^{-x} = -\frac{1}{k},$$ on solving for $x$ we have $$x=−\text{W}_\nu \left (−\frac{1}{k} \right ).$$ Here $\nu$ denotes the branch of the Lambert W function.

As $k > 0$, two distinct real roots occur when $0 < k < e$. These roots are given by $−\text{W}_0(−1/k)$ and $−\text{W}_{−1}(−1/k)$. Here $\text{W}_0(x)$ denotes the principal branch of the Lambert W function while $\text{W}_{−1}(x)$ denotes its secondary real branch.

For one real root only, this occurs at the branch point between the two real branches $\text{W}_0 (x)$ and $\text{W}_{-1} (x)$, it occurring when the argument for the Lambert W function is equal to $-1/e$. Thus $k = e$ and as $$\text{W}_0 \left (-\frac{1}{e} \right ) = \text{W}_{-1} \left (-\frac{1}{e} \right ) = -1,$$ we find it occurs at the point $x = 1$.

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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)

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