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moongazer
 4 years ago
Find the value of x and y that will satisfy the system
moongazer
 4 years ago
Find the value of x and y that will satisfy the system

This Question is Closed

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1e^x  y = 2 e^x + y = 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is xy and x+y both exponents

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349690570593:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is that the question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0introduce In both sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If that is the question then it has no real roots.

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1(e^x)  y = 2 (e^x) + y = 1

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1sorry for the late reply

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then Add equation one and two ........

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think no real solution... u might get5 imaginary....

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0still e^x = 1/2 has no real solution for x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes still no real solution

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0u have choices/options ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But that will not be true.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0imaginary solution is how?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349691346340:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349691384296:dw

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1that's also what I got

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1Isn't it that e^(ln 1/2) =1/2 ?

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1@hartnn @sauravshakya

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0yes, thats correct but y did u caculate that ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0i meant yes, thats correct but why did u caculate that ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0e^(ln 1/2) =1/2 this is hell wrong... !! the equation is not defined...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0yes, there are no real solutions to this.....

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1What should I write on my paper?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0i think u just write e^x=1/2, y =3/2 and leave it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u can do the above thing or write no real solution...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0*and write no real solution

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1I think I got the answer

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1ln of a negativ number  how it is possible ?  little up check it please !!!

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1I got x = ln 1 and y = 1

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1I think that is the answer

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1check where is defined the logarithm funvtion please

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1e^x  y = 2 e^x + y = 1

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0ln(1), 1 does not satisfy 2nd equation ln(1) is imaginary = i\(\pi\)

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1ln(1) how do you mean this ?

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1it should be ln 1/2 =x and 3/2=y

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1when by definition logb X so x need being allways positiv

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0\(e^{i\pi}=1\) by euler's relation

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1e^x  y = 2 e^x + y = 1 so and for these equations ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.0but if x is imaginary, y must be imaginary also, right ?

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1but in this case i think that is very simple us ln to both sides for every two equations and will get the right answers

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1when you substitute x with ln 1/2 it will be e^(ln 1/2)  y = 2 then 1/2y=2 y=3/2

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1I think my teacher doesn't need to know the decimal value of x.

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1ok, thanks for the answers :)

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1why is it ln (negative number) is undefined?

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1by domain definition of logarithms

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1you have learned it sure  i think ,yes ?

jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1good so than all is understood it right good luck bye

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1Thanks :) do you know how to decompose a function?

moongazer
 4 years ago
Best ResponseYou've already chosen the best response.1partial fractions the one with cases. I got an odd looking expression
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