# Where did I go wrong with my odd proof that $\frac{3dx}{3x} = \frac{5dx}{5x} \iff 3=5$?

I don't know where I went wrong, but it's interesting for me. Please check where my fault is! It is obvious that the below equation is correct:

$$\frac{3dx}{3x}=\frac{5dx}{5x}$$ $$u=3x$$and$$v=5x$$ $$\frac{du}{u}=\frac{dv}{v}$$ integrate both sides: $$ln(u)=ln(v)$$ $$u=v$$ $$3x=5x$$ so, $$3=5$$

• When you integrate both sides you need your constant of integration. "+C" – Doug M Aug 10 '18 at 20:25
• Forgot the constant of integration – Niall Aug 10 '18 at 20:25
• An excellent example of the importance of remembering the constant of integration. I will certainly use this as an example when I teach integration .... – Bruce Aug 10 '18 at 20:27
• 3 does equal 5. ... plus a constant. – fleablood Aug 10 '18 at 20:28
• @fleablood In this case multiplied by a constant. – Serge Seredenko Aug 10 '18 at 21:32

Integrating we obtain

$$\ln(u)=\ln(v)+ C$$

• Where $C = \ln u - \ln v = \ln \frac{u}{v}$ So, $\ln u = \ln \left(v \cdot \frac{u}{v}\right)$, giving us $3 = 5 \cdot \frac{3}{5}$..Spelling out what you were implying. – Davislor Aug 11 '18 at 1:16
• @Davislor I'm not implying nothing! I just gave the answer to the OP that is "I don't know where I went wrong". You are free to add your own answer here below with all the details you like. Morover try to be polite when you formulate any kind of request. – gimusi Aug 11 '18 at 6:38
• Sorry to have sounded impolite! I just wanted to acknowledge that you were aware of the steps you left out. – Davislor Aug 11 '18 at 7:22
• @Davislor Thanks for the clarification. I would like to proceed as in the answer here below exponentiating both sides to obtain $u=kv$ wiith $k=e^C=3/5$. Let add your own answer if you like to suggest another way. Bye – gimusi Aug 11 '18 at 7:33

If you're taking an indefinite integral, then you need to include constants. Integrating, we get $\ln(u)+C_1 = ln(v)+C_2$. We can collect $C_1$ and $C_2$ into one constant by setting $C_3 = C_2-C_1$, getting $\ln(u) = ln(v)+C_3$. We can then exponentiate both sides, getting $u=ve^{C_3}$. Setting $C_4=e^{C_3}$, this becomes $u=C_4v$. In this case, $C_4$ is $\frac35$.

If you're taking definite integrals, you need to include the limits. Suppose we integrate starting from $x = 1$. Then $u$ starts at $3$. So when we integrate $\frac{du}u$, we have $\ln(u)$ at the upper limit, but we have to subtract $\ln$ of the lower limit, and the lower limit is $3$. So we have that the definite integral is $\ln(u)-\ln(3)=\ln(\frac u3)$. Similarly, for $v$, we get that the definite integral is $\ln(\frac v 5)$. Plugging in $u=3x$ and $v=5x$, we get $\ln(\frac {3x}3)=\ln(\frac {5x} 5)$, which simplifies to $\ln(x)=\ln(x)$.

• After setting $C_3$, the LHS is just $ln(u)$ – CDspace Aug 10 '18 at 21:33

To turn my comment into an answer, your proof is correct up to the step $$\frac{\mathrm{d} u}{u} = \frac{\mathrm{d} v}{v}$$

At that point, as others have said, you forget to add the constants of integration when you integrate:

$$\ln u + \mathrm{C_1} = \ln v + \mathrm{C_2}$$

Substituting $\mathrm{C} = \mathrm{C}_2 - \mathrm{C}_1$ gives us:

\begin{align} \ln u &= \ln v + \mathrm{C} \\ \ln u - \ln v &= \mathrm{C} \\ \end{align}

Substituting for $\mathrm{C}$ in the first equation just gets us $\ln u = \ln u$, which is clearly no contradiction. A little more interesting, perhaps, is to turn our identity for $\mathrm{C}$, $\ln u - \ln v$, into $\ln \frac{u}{v}$ and then substitute that into the other equation above to get:

\begin{align} \ln u &= \ln v + \ln \frac{u}{v} \\ &= \ln \left( v \cdot \frac{u}{v} \right) \end{align}

Now take the exponential of both sides.

\begin{align} u &= v \cdot \frac{u}{v} \\ 3x &= 5x \cdot \frac{3x}{5x} \\ 3 &= 5 \cdot \frac{3}{5} \end{align}

This is presumably what Serge Seredenko meant when he said in his comment that 3 is equal to 5, “[i]n this case multiplied by a constant.”