Differentiating both sides of a non-differential equation I'm working on solving for $t$ in the expression $$\ln t=3\left(1-\frac{1}{t}\right)$$ and although I can easily tell by inspection and by graphing that $t=1$, I'd like to prove it more rigorously.
I got stuck trying to solve this algebraically, so I tried to take the derivative of each side with respect to $t$ to get
$$\frac{1}{t}=3\left(\frac{1}{t^{2}}\right).$$
However, this implies that $t=3$, which is incorrect.
Why can't I take the derivative of each side like this? What am I doing wrong or misunderstanding?
 A: Let $f(t) = \ln t - 3 (1 - 1/t)$. You're saying that $f(1)= 0$ and $f'(3) = 0$; there is no problem here. The point is that when you try to find a value of $t$ that satisfies the first equation, you're looking for a root of $f$, while looking for a value of $t$ that satisfies the second equation is the same as finding a root of $f'$. In general, these are not the same and you have no reason to expect that they'll give you the same answer.
I'm not sure that there are easier ways to solve this than by inspection, though.
A: To answer the general question of "Why can't [one] differentiate each side [of an equation]?": Your original equation, of the form $f(t) = g(t)$, acts as a condition (i.e., is only true for some real $t$, in this case finitely many), not as an identity (true for all $t$ in some open interval).
When you differentiate a function $f$ at one point $a$, you implicitly use the values of $f$ in some neighborhood of $a$. Since your $f$ and $g$ are not equal in any open neighborhood, you can't expect differentiating to yield a new true condition.
In case an example clarifies, take $f(t) = t$ and $g(t) = 0$. The equation $t = 0$ certainly has a solution, but differentiating both sides gives $1 = 0$.
By contrast, it's safe to differentiate both sides of, e.g., $\cos(2t) = \cos^2 t - \sin^2 t$, since this equation is true for all real $t$.
A: The derivative of a function measures the gradient of the function at various points, and this does not give you the solution for $t$ of the original.  We can use the method of differentiation to analyze the equation, however:
$$\text{Let }f(t)=\ln t-3+\frac 3t$$
$$\text{Then }f'(t)=\frac 1t-\frac 3{t^2}$$
We can see that $f'(3)=0,\forall t\in (0,3),f(t)$ is decreasing, and $\forall t\in (3,+\infty),f(t)$ is increasing, so $t=3$ is a global minimum for the valid range of $t\in \mathbb R^+.$
Continuing the analysis:
$$\ln t=3\left(1-\frac 1t\right)\implies t={e^3\over e^{\frac 3t}}\implies e^{\frac 3t}={e^3\over t}$$
which is a nice near-symmetric equation.  Observation now immediately shows that $t=1$ is a solution, but the other solution starts at the beginning again:
$$\ln t=3\left(1-\frac 1t\right)\implies \ln t^{\frac 13}=1-\frac 1t\le \ln e$$
$$\implies t^{\frac 13}\le e\implies t\le e^3\text{ and } \ln e^3\gt 3\left(1-\frac 1{e^3}\right)$$
$$\text{Also, }\ln (e^3-e^2)=2+\ln (e-1)\lt 3-\frac 3{e^3-e^2}$$
With this, we know that the other solution is for some $t\in (e^3-e^2,e^3).$  Graphing shows the actual value at $t\approx 16.8,$ and it is likely that further analysis could come up with a decent closed expression for this value.
