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Help with Algebra homework...

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platinumike

Platinum Member
Im like about to cry theese problems are so hard for me. Im having trouble on plenty of them, some i can see them making sense. This one has left me dumbfounded, we are solving literal equations, all i know is that i must take the ln(natural log) of e to bring the exponent down so i can solve for T. here is the problem:

this is how it is written:

....E
I=---....(1-e^-RT/2)
....R


you must solve for T... I know its kinda hard to write theese problems on the pc. Please help though, my book is not helping at all.
 
I am teh suck at math and cannot help you. But I can bump this for you, in the hopes that someone who can help will help.
 
you just move everything that's not e^whatever to one side then take ln of both sides and simplify
 
Originally posted by: akubi
you just move everything that's not e^whatever to one side then take ln of both sides and simplify
Click to expand...

I=E/R(1-E^-RT/2)

So it should look like this ?:

ln I-E/R-1= -RT/2 ln -E
 
No, don't waste time and effort distributing the E/R first. You're going to have to divide it out sooner or later, distributing it first just adds a step (and makes you slightly less efficient)

I can't quite tell what you mean by the E/R(1-...)
Is E/R a coefficient written in front of the (1-...) or is what's in parentheses in the denominator?

Try this: cross multiply first. (handles both situations I have above)
 

I=E/R(1-E^-RT/2)

is it

...........E
I = ------------
.......R(1-E^RT/2)


or is it

I = E(1-E^RT/2)
.....--------------
.............R
 
Originally posted by: DocHolliday
so glad i dont' have to look at that sort of thing anymore...
Click to expand...

I got a degree in computer networking and I NEVER looked at anything like that.
 
If it's the latter and you keep in mind "cross-multiplying" (which students seem to understand better, even though it's not really that)

Then, IR = E(...)
IR/E = 1 - E^...
IR/E -1 = (IR-E)/E = -E^...
get rid of the negative
(E-IR)/E = E^-RT/2

And for God's sake, if this capital E is supposed to be Euler's number (pronounce "oiler's number"), then use a lower case e!!!

For what it's worth, when you ln the left (assuming E is really e), use properties of logs to break the natural log up, and lne = 1.
 
Originally posted by: DrPizza
No, don't waste time and effort distributing the E/R first. You're going to have to divide it out sooner or later, distributing it first just adds a step (and makes you slightly less efficient)

I can't quite tell what you mean by the E/R(1-...)
Is E/R a coefficient written in front of the (1-...) or is what's in parentheses in the denominator?

Try this: cross multiply first. (handles both situations I have above)
Click to expand...

youre right, multiply R/E to both sides then add one, depending if its E/R(1-...) or E(1-...)/R. btw OP, is it E^-(RT/2) or E^-r*(T/2) or (E^-(rt))/2 or what? its kinda confusing
 
this is how it is written:

....E
I=---....(1-e^-RT/2)
....R

the e in the parantheses is a lower case e, sorry my mistake. The one in the fraction is written in my book as a capital E
 
THe actual answer in the book is

T=-2/Rln(1-RI/E)

xcobra you are really really close i think you just misplaced the r and forgot the 1
 
you didnt say whether it was E^-(RT/2) or E^-r*(T/2) or (E^-(rt))/2, or whether E^-(..) was a denominator or numerator..
 
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