# How to solve $\ln t = t-1$ to get $t=1$?

How can one solve the equation $$\ln t = t-1$$ to get $$t=1$$?

It is clear the answer is one but how can we solve this to get $$t=1$$ and remove the $$\ln$$?

• Graphing each side and doing a guess and check would be a quick approach to see if there is an easy answer... Mar 13, 2019 at 4:08
• @abiessu: I know the answer, I want to show it rigorously. It is clear the answer is one but how can you solve this to get $t=1$? Mar 13, 2019 at 4:11

You might "guess" that $$1$$ is a solution (and then wonder if there are others). If $$1$$ doesn't stand out as a self evident solution, then using the Lambert W function, we have

\begin{align} \ln(t)&=t-1\\ t&=e^{t-1}\\ te^{-t}&=e^{-1}\\ -te^{-t}&=-e^{-1}\\ W\mathopen{}\left(-te^{-t}\right)\mathclose{}&=W\mathopen{}\left(-e^{-1}\right)\mathclose{}\\ -t&=W\mathopen{}\left(-e^{-1}\right)\mathclose{}\\ t&=-W\mathopen{}\left(-e^{-1}\right)\mathclose{}\\ \end{align}

Using known special values of this function, we conclude $$t=1$$.

Understanding this function, we might worry about additional solutions from other branches of the W function. However, now that we know $$t=1$$ is a solution (which again, maybe you could have seen without Lambert), we can prove it is the only solution.

$$\ln(t)$$ is concave down, and $$t-1$$ is linear. So either there are two intersections, or there is one intersection and at that intersection the two curves have the same derivative. In this case, the latter option holds true. We check that $$\left.\frac{d}{dt}\ln(t)\right|_{t=1}=\left.\frac{1}{t}\right|_{t=1}=1$$, and $$\left.\frac{d}{dt}(t-1)\right|_{t=1}=\left.1\right|_{t=1}=1$$, so they are equal. The solution at $$t=1$$ is a solution with tangency, and therefore the only solution.

Several answers have been given, but I don't think any have addressed the question as you present it. It seems you want a series of algebraic steps by which to manipulate the statement $$\ln(t)=t-1$$ into the statement $$t=1$$, and the correct answer to this request is that no such sequence exists, at least in terms of elementary operations (as seen in alex.jordan's answer, this claim is not true without the caveat).

You can always rewrite an equation of a single real variable into the form $$F(t)=c$$, where $$F: A \to \mathbb{R}$$ is some map and $$A$$ is some subset of $$\mathbb{R}$$, and these equations can typically be solved "nicely" when the equation can be algebraically cast in such a way that $$F$$ is "nice", which in this context would mean has an inverse expressible in terms of well-known functions-- the exponential map, logarithm, polynomial functions, trigonometric functions, etc.-- at least in a neighborhood of the solutions. The problem is that the vast majority of equations you can write down are not "nice" in this way, and $$F(t)=\ln(t)-t+1=0$$ is one such example.

In cases where $$F$$ is not "nice", it is often all you can do to observe solutions of the equation by direct computation, as multiple answers have done, and if you're lucky you can prove uniqueness. In our example, we are lucky in this way: since $$F'(t)=\frac{1}{t}-1$$, we see that $$F$$ achieves a global maximum at $$t=1$$, so $$F$$ can never attain $$0$$ at a point other than $$t=1$$.

• With the Lambert function (which is not elementary) you can have such a sequence of algebra manipulations. Mar 13, 2019 at 5:20
• Ah, I see! I was not familiar with this function. Thanks for that tip! I think I stand by my overall treatment of the question, given the "non-elementariness" of the $W$ function, but it's nice to see the formal solution in the OP's sense indeed exists. Mar 13, 2019 at 5:32
• @alex.jordan I think it worth stating, though, that the requirement of "elementary" steps and functions is not at all unreasonable. Indeed, any equation with $F$ as above injective on some neighborhood of the solutions (say, $\exp(\sin(t))+\cos(t) = 2$) can be formally solved in the sense of the OP by simply defining the local inverse to $F$ and writing $t = F^{-1}(c)$, precisely as the $W$ function does, but this often helps no more than the statement $F(t)=c$ unless $F^{-1}$ is understandable in terms of familiar operations. Mar 13, 2019 at 6:08
• I noticed that every single answer except yours has a downvote. Mar 27, 2019 at 22:30
• @ChaseRyanTaylor I'm not sure if you're insinuating that I downvoted the other answers to improve the reception of my own, but I assure that none of those downvotes are due to me. The only feedback I've given on other answers is an upvote on alex.jordan's. Mar 28, 2019 at 0:15