Question

ln (y+6)- ln(4)= x+ ln(x); solve for y! ? (: Please explain!

Answer 1

\[\Large \ln(y+6)-\ln4=x+\ln x\]

Answer 2

Hmm ok so we'll start by applying a rule of logs,\[\large \color{royalblue}{\log(a)-\log(b)=\log\left(\frac{a}{b}\right)}\]
Do you see what that will do to our left side? :)

Answer 3

divide? (:

Answer 4

Yes! We can combine the logs and divide.\[\Large \ln\left(\frac{y+6}{4}\right)=x+\ln x\]

Answer 5

ok ...(:

Answer 6

This next step is a little tricky.
We want to `exponentiate` both sides.

We'll rewrite both sides as exponents with a base of e.\[\Large e^{\ln\left(\frac{y+6}{4}\right)}=e^{(x+\ln x)}\]

Answer 7

The `log` and the `exponential` functions are `inverses` of one another.
So they will essentially "undo" one another.

The left simplifies to,\[\Large \frac{y+6}{4}\]

Answer 8

So that gives us,\[\Large \frac{y+6}{4}=e^{(x+\ln x)}\]Which we can simplify a bit further.

Answer 9

I know I know, the exponential thing was a bit confusing right? D:
Let's do the rest of the steps though.
Maybe we can go back to that in a moment :3

Answer 10

okay :)

Answer 11

do the e and ln cancel on the right side?

Answer 12

Yes very good.
But we need to fuss with it a bit first.

Recall a rule of exponents:\[\large \color{royalblue}{e^{a+b}=e^a\cdot e^b}\]
See how we can apply this rule to the right side of our equation? :o