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View Full Version : [Logarithms] help me please?


ShadowOp
04-21-2008, 01:53 PM
can someone help me work out

solve for x
e^x-1 = 9-18e^1-x


and



w/o a calculator, solve for x and y
3^4x = 81(1/9^1-y)
xln8 - yln2 = ln64




oh yeah provide me with the solutions too. Thanks!!

boinker
04-21-2008, 01:56 PM
onrpg isnt a math calculator , we r just dumbz and you suspect us to do that simple equation D:

ShadowOp
04-21-2008, 02:02 PM
come on just help me out a bit >.< i need it for my homework

boinker
04-21-2008, 02:10 PM
ill try it in a little bit >.<

TaintedEon
04-21-2008, 03:00 PM
I believe you set E = 0, and then solve, and then set X = 0, and then solve.

Raiyne
04-21-2008, 04:02 PM
No, e has a set value of 2.something, sort of like Pi.

Zynosphere
04-21-2008, 04:37 PM
o.o what the hell is that?

Xegonyv
04-21-2008, 11:04 PM
Lol, multiply ln to cancel out each of the e's on both sides, get the variable on one side, then use the calculator to solve for x.

That's just the first prob. I'm too lazy to do the others :P Btw, are you taking Calculus or Trigonometry?

Edit: Worked out number 1, not sure if I'm right.

e^x-1 = 9-18e^1-x
then: e^(x-1)-9 = -18e^(1-x)
then: ln(e^(x-1))-9 = ln(-18e^(1-x))
ln and e cancel out: (x-1)-9 = -18(1-x)
distributive property: x-10 = -18-18x
algebra 1: 19x = -8
then: x = -8/19

Hope that helps. Plug it into your calculator and see if both sides match :|

Mr. First Name Basis
04-21-2008, 11:17 PM
Lol, multiply ln to cancel out each of the e's on both sides, get the variable on one side, then use the calculator to solve for x.

That's just the first prob. I'm too lazy to do the others :P Btw, are you taking Calculus or Trigonometry?

Edit: Worked out number 1, not sure if I'm right.

e^x-1 = 9-18e^1-x
then: e^(x-1)-9 = -18e^(1-x)
then: ln(e^(x-1))-9 = ln(-18e^(1-x))
ln and e cancel out: (x-1)-9 = -18(1-x)
distributive property: x-10 = -18+18x
algebra 1: 19x = -8
then: x = -8/19

Hope that helps. Plug it into your calculator and see if both sides match :|

Negative times a negative is a positive. Small mistake but otherwise correct from what I see.

EDIT: Actually no, to cancel out e with natural log, you must do the whole side, which in this case you left the 9 down. So yeah.. not exactly right. Right concept, hold on I'll solve by hand but by doing it through graphing means on calculator I got x = 2.79 and x = 2.09.

I'll solve it by hand in a second.

NVM I tried and i'm not getting what i got in the calculator. If you want to know how to solve it in the calc easy, if you have a TI83 and beyond, just put both sides of the equation into Y1 and Y2 (Y=) and then find the intersection. In this case, there were 2.