Why does $e^{t+4}(t-1)=0$ only yield one solution?

Question

I understand that $e^{t+4}(t-1)=0$ only yields one solution because $e^x >0$ what I am wondering is why you can't $\ln$ to cancel out the $e$ to give $(t+4)(t-1)=0$

This is clearly invalid from looking at online resources and there's also no need to do it but Maths doesn't lie after all and I'm wondering why this wouldn't be possible. The only thing I can possibly think of is that you have to $\ln 0$ on the right hand side despite it being $0$

Answer 1

Notice that $e^{t+4}(t-1) = 0$ is factored. That is you have set up a product of two terms that's equal to zero. That would imply that: $t-1 = 0$ or $e^{t+4} = 0$. For the equation, $e^{t+4} = 0$, to isolate for the exponent $t+4$, you would have to take $ln$ of both sides. But the value of $ln(0)$ is undefined. The reason it's undefined is that the domain of $ln(x)$ is $\{x \in \mathbb{R} | x > 0\}$. This is precisely a consequence of the fact that $e^x > 0$, as you point out in your question. Since $lnx$ is the inverse operation to $e^x$ the Range of $e^x$ is also the Domain of $lnx$

That all said, to answer your question about why you can't take the $ln$ of your original equation, not only would you be calculating $ln(0)$, but your algebra is off. You can't just take the $ln$ of one term in a product. In abstract, you could do: $ln(e^{t+4}(t-1)) = ln(e^{t+4}) + ln(t-1)) = t+4 + ln(t-1)$, but I'm not sure that's a very helpful algebraic manipulation. But even then, in the context of this question, if you did $ln(e^{t+4}(t-1))$, you would also have to take the $ln$ of the other side, $0$, and as mentioned, $ln(0)$ is undefined.

Answer 2

If you tried to use $\ln$ on both sides you get

$\ln [e^{t+4}(t-1)] = \ln 0$

$\ln (e^{t+4}) + \ln (t-1) = \ln 0$

$t+4 + \ln(t-1) = \ln 0$.

And from here we could not the RHS is undefined $\lim_{x\to 0^+} x=-\infty$. So we must have the LEFT also undefined. So $t-1 \le 0$. But if $t-1 < 0$ then we don't have any value for $\lim_{x\to t-1} \ln x$ not even $-\infty$. As $x$ can't be near $t-1$ if $t-1 < 0$ then no limit makes any sense.

So we must have $t-1 = 0$ and $t =1$.

So... you can almost do that.

.......

What you seem to want to do is selectively apply $\ln e^{t+4}$ but NOT apply it to $(t-1)$ and .... you just can't do that.

You certainly can NOT say $[e^{t+4} ](t-1) = 0$ so $[\ln e^{t+4}](t-1) = 0$. That's just nutty. $A\ne \ln A$ so it make no sense to say $AB = (\ln A)B$ but that's exactly what you are trying to say.