in(1-3x)+3=9
Is it: \(ln(1-3x)+3=9\)?
yes
\(log_e(1-3x)=6\) => \((1-3x) = e^{6}\) Are you able to do it now?
no :/
1-3x=e^6, try solving that for x
\(-3x = e^{6} - 1 \)
well, looks like you were left in the cold. from here you divide both sides by -3 to get \[x=\frac{e^6-1}{-3}\]
\[e^6 = 403.428793\]
now minus 1 and then divide by -3
how do you guys know this stuff thank u
its my career and hobby. teacher and mathematician. don't be discouraged! you can learn it too. here are some awesome places to check out to help coolmath.com khanacademy.org patrickjmt.com that's just the tip of the iceberg. many free and useful resources out there for you to learn. see ya.
thank you very much i appreciate that i often feel discouraged by math and just want to give up because for the past 3 years i just seem to get worse and worse at it and i have to finish this summer class in order to 2 pass this summer class ugghhh and do you know any sites for logarithms and equating exponents and things of that nature?
http://coolmath.com/algebra/17-exponentials-logarithms/index.html plus check out the rest of the algebra section. it will help if you search out video lessons and notes and practice problems.
you might want to go right to "what is a logarithm?"
thank you i need as much help conquering this as possible because its just going to get worse so im trying to get a handle on it before its just of hand and besides i hate being dominated by things :D
the more you do and the more you try to practice the better it will get. start saying "it's going to get better if I keep working hard and put in time to practice everyday." that's what I say about everything from teaching to learning to parenting to life...
thank you very much this means alot to me :) god bless
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