Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

sure

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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