# Redefinition of a Constant Leading to Nullification of Absolute Value

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.

• according to the property of exponentiation: $e^{2Cx}=(e^{2C})^x=(e^C)^{2x}=(e^{2x})^C\ne e^{2x}e^C=e^{2x+C}$. – farruhota Apr 5 at 19:27
• 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! – Little Boy Blue Apr 5 at 21:00
• 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! – Little Boy Blue Apr 5 at 21:06

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.
• 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. – Siong Thye Goh Apr 6 at 2:58