Solve each equation. Problem: ln5-ln-4x=4

\[x=-\frac{5}{4e^{4}}\]

How did you get that? Can you draw it for me? Thanks so much for helping me!

\[\ln(5)-\ln(-4x)=4\] \[-\ln(-4x)=4-\ln(5)\] \[\ln(-4x)=\ln(5)-4\] \[e^{\ln(-4x)}=e^{ln(5)-4}\] \[-4x=5*e^{-4}\] \[-4x=\frac{5}{e^4}\] \[x=\frac{-5}{4e^4}\]

Oh my gosh thanks so much! Would you mind helping me on some more problems step by step like that?

sure

Thank you so much! You dont know how happy I am right now! So next problem is ln(x+4)- lnx=4

this is pretty similar, except we can add the logs to begin with (rules of logs): \[\ln\frac{x+4}{x}=4\] \[\huge e^{\ln\frac{x+4}{x}}=e^{4}\] \[\frac{x+4}{x}=e^4\] \[\frac{x}{x}+\frac{4}{x}=e^4\] \[1+4/x=e^4\] \[4/x=e^4-1\] \[1/x=1/4(e^4-1)\] you can think of 1/x as x^-1 and then you can see you need to raise both to ^-1 since (x^-1)^-1=x \[x=(1/4(e^4-1))^{-1}\] \[x=\frac{4}{-1+e^4}\] the last step is just switching back to fractions.

Okay thanks. But wouldnt \[e^4 \] just be 4?

no, e is a constant e=2.71828 so e^4 is more on the order of 54.5something

the reason the logs cancel is because ln is log base e so taking it to the e base causes ln to cancel. e^(lnx)=x

Okay thats starting to make more sense. Should i do another problem for you and see what I could do so far?

Unless your busy? Thanks so much for the help though. I REALLY appreciate it.

go for it, im just watching family guy and messing around on here right now :P

Haha oh my gosh family guy. Okay well ill start on another problem. Hold on.

|dw:1319600054479:dw| Okay so I got here so far. But I was thinking that isnt it only for minus that you do the division?

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