Express y as a function of x if ln(y-3)= -4x + ln5.

[Red]

Express y as a function of x if ln(y-3)= -4x + ln5.

I moved 4x to the left side of equation and ln(y-3) to the right, giving me
4x=ln5 - ln(y-3) Do I now use the quotient property of logs?? x=ln5/ln(y-3)4??

 

[galactus]

You have:

\(\displaystyle \L\\ln(y-3)=-4x+ln(5)\)

Take e to both sides:

\(\displaystyle \L\\e^{ln(y-3)}=e^{-4x+ln(5)}\)

\(\displaystyle \L\\y-3=e^{-4x+ln(5)}\)

\(\displaystyle \L\\y=5e^{-4x}+3\)

 

[Red]

I guess I'm not understanding this answer because this problem fell on a quiz covering the properties of logarithms. This question was supposed to cover combining the terms. Thank you anyway

 

[galactus]

Well, we could try something else. Maybe they want something like this:

Subtract ln(5) from both sides:

\(\displaystyle \L\\ln(y-3)-ln(5)=ln(\frac{y-3}{5})\)

Then you have:

\(\displaystyle \L\\ln(\frac{y-3}{5})=-4x\)