anonymous
  • anonymous
ln (y+6)- ln(4)= x+ ln(x); solve for y! ? (: Please explain!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
\[\Large \ln(y+6)-\ln4=x+\ln x\]
zepdrix
  • zepdrix
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? :)
anonymous
  • anonymous
divide? (:

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zepdrix
  • zepdrix
Yes! We can combine the logs and divide.\[\Large \ln\left(\frac{y+6}{4}\right)=x+\ln x\]
anonymous
  • anonymous
ok ...(:
zepdrix
  • zepdrix
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.\[\Large e^{\ln\left(\frac{y+6}{4}\right)}=e^{(x+\ln x)}\]
zepdrix
  • zepdrix
The `log` and the `exponential` functions are `inverses` of one another. So they will essentially "undo" one another. The left simplifies to,\[\Large \frac{y+6}{4}\]
zepdrix
  • zepdrix
So that gives us,\[\Large \frac{y+6}{4}=e^{(x+\ln x)}\]Which we can simplify a bit further.
zepdrix
  • zepdrix
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
anonymous
  • anonymous
okay :)
anonymous
  • anonymous
do the e and ln cancel on the right side?
zepdrix
  • zepdrix
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

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