If lny + lnx = lnc, how do you find an expression for y?

`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`.