Can anyone explain this property related to logs?

Question

We have $$\ln(v-1) - \ln(v+3) = \ln(x) + C$$

Multiplying through by e gives:

$$(v-1)/(v+3) = x + e^C$$

But the answer in the textbook is:

$$(v-1)/(v+3) = Dx$$

Where $$D = e^C$$

Question: why is the answer Dx and not x + D?

Thanks!

Answer 1

Because $$ e^{\ln(x)+C} = e^{\ln(x)} e^C = x e^C. $$

Answer 2

You're not multiplying through by $e$. You're exponentiating both sides of the equation. So your second equation should be:

$$\frac{v-1}{v+3} = e^Cx.$$

Answer 3

$ln(\frac{v-1}{v+3})$ $=$ $ln(x)+C$ $\implies$ $\frac{v-1}{v+3} = Dx$.