Mathematics OpenStudy (anonymous):

Solve each equation. Problem: ln5-ln-4x=4 OpenStudy (agreene):

$x=-\frac{5}{4e^{4}}$ OpenStudy (anonymous):

How did you get that? Can you draw it for me? Thanks so much for helping me! OpenStudy (agreene):

$\ln(5)-\ln(-4x)=4$ $-\ln(-4x)=4-\ln(5)$ $\ln(-4x)=\ln(5)-4$ $e^{\ln(-4x)}=e^{ln(5)-4}$ $-4x=5*e^{-4}$ $-4x=\frac{5}{e^4}$ $x=\frac{-5}{4e^4}$ OpenStudy (anonymous):

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

sure OpenStudy (anonymous):

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

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. OpenStudy (anonymous):

Okay thanks. But wouldnt $e^4$ just be 4? OpenStudy (agreene):

no, e is a constant e=2.71828 so e^4 is more on the order of 54.5something OpenStudy (agreene):

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 OpenStudy (anonymous):

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

Unless your busy? Thanks so much for the help though. I REALLY appreciate it. OpenStudy (agreene):

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

Haha oh my gosh family guy. Okay well ill start on another problem. Hold on. OpenStudy (anonymous):

|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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