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**Soroban** Hello, Ashley!

You should be familiar with this procedure by now . . .

First of all, I used two different C's.

I had an arbitrary constant, $\displaystyle c.$

Since $\displaystyle c$ is the log of some other number, $\displaystyle C$,

. . I let: .$\displaystyle c \:=\:\ln(C)$

Now why do we need something like that?

Suppose our equation came out to: .$\displaystyle \ln y \:=\:2\ln x + c$

We can exponentiate both sides: .$\displaystyle e^{\ln y} \;=\;e^{2\ln x + c}$

. . and get: .$\displaystyle y \;=\;e^{2\ln x + c}$

Technically, we're done . . . We have the answer.

But we should not be surprised that the book has *simplified* the answer.

Before we exponentiate, we have:

. . $\displaystyle \ln y \;=\; \underbrace{2\ln x}_{\downarrow} + \underbrace{c}_{\downarrow}$

. . $\displaystyle \ln y= \;\underbrace{\ln x^2 + \ln C}_{\downarrow}$

. . $\displaystyle \ln y \;=\;\;\;\overbrace{\ln(Cx^2)}$

Now we can exponentiate: .$\displaystyle y \;=\;Cx^2$ . . . see?