1. ## [SOLVED] natural log

i have this equation:

1=10*(e^x)-(e^y)

and i need to find y-x, but i don't think i can just take the natural log of both sides?

2. Originally Posted by junior
i have this equation:

1=10*(e^x)-(e^y)

and i need to find y-x, but i don't think i can just take the natural log of both sides?
to clarify, do you mean $1 = 10e^x - e^y$? and was this the original problem?

3. Yes.

wait i think i got it:

ln(1) = ln(10)+x-y ?

4. Originally Posted by junior
Yes.

wait i think i got it:

ln(1) = ln(10)+x-y ?
actually no. you cannot distribute a log across a sum...

5. so this cant be simplified??

6. Originally Posted by junior
so this cant be simplified??
to find y - x in terms other than y and/or x? no, not anyway that i am aware of. at least, not using any kind of pre-university math... which is why i asked if this was how the problem was presented originally. but you said yes. i don't see a way to get to the answer.

7. maybe you can still help me:

im basically trying to solve number 6, so but im just doing the math wrong. the solution posted here does the problem by dividing, i tried subtracting the equations instead: http://www.ece.msstate.edu/~ykoshka/...Solutions).pdf

$
I = I_s *(e^(V_1)/0.026)-1)$
(1)

$10 I = I_s*(e^(V_2/.026)-1)$ (2)

subtracting (2)-(1) i end up with this:

$10=[e^(V2/.026)-1]/(e^(v1/.026)-1)$

how do i solve this?

8. actually nvm, i think i made a critical error

that equation is hard to solve, unless you approximate

$
e^v-1=e^v
$

9. Originally Posted by junior
maybe you can still help me:

im basically trying to solve number 6, so but im just doing the math wrong. the solution posted here does the problem by dividing, i tried subtracting the equations instead: http://www.ece.msstate.edu/~ykoshka/ECE4243-6243-Spring2008/HW/HW8(Solutions).pdf

$
I = I_s *(e^(V_1)/0.026)-1)$
(1)

$10 I = I_s*(e^(V_2/.026)-1)$ (2)

subtracting (2)-(1) i end up with this:

$10=[e^(V2/.026)-1]/(e^(v1/.026)-1)$

how do i solve this?
why don't you want to use division? it seems a lot easier to get to what you want to get to. also note that the solutions do not use the original equation, but an approximation of it. it would stil be really difficult to solve the equation as you have it. in any case, this looks like physics, and i'm not the expert on that.

10. Originally Posted by junior
actually nvm, i think i made a critical error

that equation is hard to solve, unless you approximate

$
e^v-1=e^v
$
right. apparently $I_s$ is really small