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e^x - y = -2
e^x + y = 1

is x-y and x+y both exponents

|dw:1349690570593:dw|

Is that the question?

introduce In both sides

If that is the question then it has no real roots.

how do I solve it?

(e^x) - y = -2
(e^x) + y = 1

sorry for the late reply

Then Add equation one and two ........

i think no real solution...
u might get5 imaginary....

still e^x = -1/2 has no real solution for x

Yes still no real solution

y comes out to be 3/2

u have choices/options ?

But that will not be true.

I don't have

imaginary solution is how?

|dw:1349691346340:dw|

|dw:1349691384296:dw|

that's also what I got

http://www.wolframalpha.com/input/?i=e%5Ex+%3D+-1%2F2

ok

Isn't it that e^(ln -1/2) =-1/2 ?

yes, thats correct but y did u caculate that ?

nope

i meant yes, thats correct but why did u caculate that ?

e^(ln -1/2) =-1/2 this is hell wrong... !!
the equation is not defined...

to find y

I got y=3/2

yes, there are no real solutions to this.....

impossible man..

What should I write on my paper?

i think u just write e^x=-1/2, y =3/2 and leave it

u can do the above thing or write no real solution...

*and write no real solution

I think I got the answer

ln of a negativ number - how it is possible ? --- little up check it please !!!

I got x = ln -1
and y = 1

I think that is the answer

check where is defined the logarithm funvtion please

function --- sorry

I forgot to add e^x

e^x - y = -2
e^x + y = 1

ln(-1), 1 does not satisfy 2nd equation
ln(-1) is imaginary = i\(\pi\)

ln(-1) how do you mean this ?

it should be ln -1/2 =x
and 3/2=y

when by definition logb X so x need being allways positiv

\(e^{i\pi}=-1\)
by euler's relation

e^x - y = -2
e^x + y = 1
so and for these equations ?

but if x is imaginary, y must be imaginary also, right ?

when you substitute x with ln -1/2
it will be
e^(ln -1/2) - y = -2
then
-1/2-y=-2
y=3/2

I think my teacher doesn't need to know the decimal value of x.

ok, then write that....

ok, thanks for the answers :)

why is it ln (negative number) is undefined?

by domain definition of logarithms

you have learned it sure --- i think ,yes ?

ohh yes

because of the graph

good so than all is understood it right
good luck
bye

Thanks :)
do you know how to decompose a function?

how do you think it ?

partial fractions the one with cases.
I got an odd looking expression

i'll post it