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

Jul 10, 2016

We can manipulate the equation to get $\frac{c}{x} = y$.
An expression for $y$ is therefore $\frac{c}{x}$.

Explanation:

$\ln y + \ln x = \ln c$
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:

$\textcolor{g r e e n}{\ln a - \ln b = \ln \left(\frac{a}{b}\right)}$ for any $a$ and $b$.

So, we have:
$\ln y = \ln c - \ln x$

$\setminus R i g h t a r r o w \ln y = \ln \left(\frac{c}{x}\right)$

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

Solution 1:

Remember that:
$\textcolor{g r e e n}{a = \ln b}$ means that $\textcolor{g r e e n}{{e}^{a} = b}$.

So, we can use this to get:

$\ln y = \ln \left(\frac{c}{x}\right)$
$\setminus R i g h t a r r o w {e}^{\ln \left(\frac{c}{x}\right)} = y$.

(in this case, our $\textcolor{g r e e n}{a}$ is $\ln \left(\frac{c}{x}\right)$ and our $\textcolor{g r e e n}{b}$ is $y$).

Then, we can simplify ${e}^{\ln \left(\frac{c}{x}\right)} = y$ using the following rule:
$\textcolor{g r e e n}{{e}^{\ln} \left(a\right) = a}$ for any $\textcolor{g r e e n}{a}$.

So we have:
${e}^{\ln \left(\frac{c}{x}\right)} = y$
$\setminus R i g h t a r r o w \frac{c}{x} = y$.

Solution 2:
Another way to solve for $y$ using $\ln y = \ln \left(\frac{c}{x}\right)$ 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.

$\ln y = \ln \left(\frac{c}{x}\right)$
$\setminus R i g h t a r r o w {e}^{\ln y} = {e}^{\ln \left(\frac{c}{x}\right)}$

Then we can use the following rule:
$\textcolor{g r e e n}{{e}^{\ln} \left(a\right) = a}$ for any $\textcolor{g r e e n}{a}$.

${e}^{\ln y} = {e}^{\ln \left(\frac{c}{x}\right)}$
$\setminus R i g h t a r r o w y = \frac{c}{x}$.