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I am currently taking a calculus 3 class in college, and my teacher did an intriguing problem about a tsunami in class which took up about 4 full white boards. Anyways, at a certain point in the problem, we arrived at $ |{u-2 \over u+2 }| = e^{2Cx}$

Now, of course we can say that this is the same as $ |{u-2 \over u+2 }| = e^{2x}e^C$, but this is where things get a little funky. Since $e^c$ is just another constant, we can rewrite the equation as $ |{u-2 \over u+2 }| = e^{2x}C$

Alright, so eventually we got the answer $ C= -1$. This did indeed solve our problem, but then I noticed the serious issue that occurs when we go back and apply this to our equation. At the point $ |{u-2 \over u+2 }| = e^{2Cx}$, $C = -1$ is perfectly acceptable, since $e^x$ can never be negative. But our issue comes about when we redefine $e^C$ as a new constant $C$. With our answer $C = -1$, we wind up getting $ |{u-2 \over u+2 }| = -e^{2x}$

Uh-oh! Clearly this will not do, since it's impossible! Evidently, the error occurred when we redefined the constant -- but why? My teacher's solution was to simply remove $|...|$ from the equation, but this seems to be an illogical and unsatisfactory answer. So then, what went wrong? Can redefining a constant lead to fatal errors in math? Thanks for your answers.

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    $\begingroup$ according to the property of exponentiation: $e^{2Cx}=(e^{2C})^x=(e^C)^{2x}=(e^{2x})^C\ne e^{2x}e^C=e^{2x+C}$. $\endgroup$ – farruhota Apr 5 at 19:27
  • $\begingroup$ Oh wow @farruhota , thank you! The answer's always simpler than you expect! Would you mind putting this as an answer, so that I can give you credit? Thanks again! $\endgroup$ – Little Boy Blue Apr 5 at 21:00
  • $\begingroup$ Actually, @farruhota , hold on there. Although what you said is correct, we can still take (e^C)^2x and define e^C as a new constant C. For C = -1, -1^some_power is still a negative number -- thus the questions still goes unanswered! $\endgroup$ – Little Boy Blue Apr 5 at 21:06
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Note that we can express $\exp(2x+C)=\exp(C)\exp(2x)=D\exp(2x)$

but we can't expressed $\exp(2Cx)$ as $D\exp(2x)$. There is indeed an error in your simplification.

You might also like to use different notation as you perform a substitution to avoid confusion.

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  • $\begingroup$ But that's the point of my question: WHY can't we? WHY did this error occur? exp(C) is a constant, after all. Thanks for the tip about renaming a new constant, by the way. $\endgroup$ – Little Boy Blue Apr 5 at 18:49
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    $\begingroup$ Suppose We express $\exp(2Cx)=D\exp(2x)$, then we have $\exp(2(C-1)x)=D$, are you claiming that $C=1$ so that $D$ is a constant? The problem with reusing a constant is that when you say $C=-1$, which $C$ are you referring to. $\endgroup$ – Siong Thye Goh Apr 6 at 2:58

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