#ln y + ln x = ln c#

We want to isolate #y#, so we move all terms not containing #y# to one side.

We can do this by subtracting #ln x# from both sides.

#ln y + ln x = ln c#

#ln y + ln x - ln x= ln c - ln x#

#ln y = ln c - ln x#

Then, we use the following law of logarithms:

#color(green)(ln a - ln b = ln (a/b)) # for any #a# and #b#.

So, we have:

#ln y = ln c - ln x#

#\Rightarrow ln y = ln (c/x)#

I'll show two ways to to solve for #y# from this step:

**Solution 1:**

Remember that:

#color(green)(a = ln b)# means that #color(green)(e^a = b)#.

So, we can use this to get:

#ln y = ln (c/x)#

#\Rightarrow e^(ln(c/x))=y#.

(in this case, our #color(green)(a)# is #ln (c/x)# and our #color(green)(b)# is #y#).

Then, we can simplify #e^(ln(c/x))=y# using the following rule:

#color(green)(e^ln(a) = a)# for any #color(green)(a)#.

So we have:

#e^(ln(c/x))=y#

#\Rightarrow c/x = y#.

**Solution 2:**

Another way to solve for #y# using #ln y = ln (c/x)# is as follows:

Just like how you can add and subtract to both sides of an equation, you can also use both sides as exponents.

#ln y = ln (c/x)#

#\Rightarrow e^(ln y) = e^(ln (c/x))#

Then we can use the following rule:

#color(green)(e^ln(a) = a)# for any #color(green)(a)#.

#e^(ln y) = e^(ln (c/x))#

#\Rightarrow y = c/x#.