Question

how do you even solve this?
1) y = 2^x ; y = x + 1
and
2) y = e^-x ; y = x^2

Answer 1

I guess my question is, what do you mean by solve?

Answer 2

(x,y) coordinates

Answer 3

well, if you just need some, just pick a few x values and plug them in where you see an x. x=0,-1,1 are usually good choices

Answer 4

ah, then that solves 1, I guess there was no proper way to solve 1.
How about 2?

Answer 5

same thing?

Answer 6

do you mean solve the system of equations ie where they are equal?

Answer 7

Yeah.

Answer 8

that's a useful bit of info

Answer 9

Sorry, just found it in the book.

Answer 10

so set \[e^{-x}=x^2\]

Answer 11

(y=y)<-- that way you can find the answer, so then, solve for x or use a quick observation

Answer 12

what is the biggest e^-x can ever be?

Answer 13

0?

Answer 14

2^0 doesn't equal zero does it?

Answer 15

and asked for the biggest it could be

Answer 16

also, another way is to just graph them and take a look at the point of intersection

Answer 17

It's 1, sorry, it was either 1 or 0 when I looked at it. Didn't think it through xD

Answer 18

*biggest it can be provided x is positive, **

Answer 19

but yea, 1, does x^2 ever equal 1?

Answer 20

yeah

Answer 21

are the x values the same when it happens?

Answer 22

no.

Answer 23

ok so the algebraic way is messy, I recommend graphing it's easier. Here use this https://www.desmos.com/calculator

Answer 24

I solved it with this but yeah, Algebraically would have been nice xD

Answer 25

\[e^{-x}=x^2\]
we'd have to use a log.... hmm -x=2lnx

Answer 26

-x/lnx=2
ln e^-x/lnx=2
ln(e^{-x}-x)=2

Answer 27

e^{-x}-x=e^2... hmm

Answer 28

hmm.... or maybe since y=e^{-x} then \[x=-lny\] and \[x=\frac{+}{-}\sqrt{x}\]

Answer 29

we get y=2lny then hmm

Answer 30

still can't solve

Answer 31

any ideas on an algebraic way guys? @ganeshie8 , @eliassaab , @hero, @ikram002p , @ wio

Answer 32

You don't have to solve that second one anymore, unless you want to. I just saw it and wanted to know how.

Answer 33

I do too... that's the thing... I can't figure it out

Answer 34

\[
y=x+1; \quad y= 2^x\\
2^x=x+1\\
x=0 \text { or } x=1
\]

Answer 35

Oh, thanks.

Answer 36

familiar with lambert w function ?

Answer 37

@Kainui

Answer 38

For the second one there no really elementary algebraic solution unless you use the productlog function. A numerical solution will give x= .7034

Answer 39

\[\LARGE e^{-x}=x^2 \]\[\LARGE 1 = x^2 e^x\]\[\LARGE \pm 1 = xe^{x/2} \]\[\LARGE \frac{\pm 1}{2}=\frac{x}{2}e^{x/2} \]\[\LARGE W(\frac{\pm1}{2})=\frac{x}{2}\]\[\LARGE 2 W(\frac{\pm 1}{2})=x\]

So although this is one answer: http://www.wolframalpha.com/input/?i=2*W(1%2F2)&t=crmtb01

the other answer is a complex number since the smallest real value is -1/e which is close to -1/3: http://www.wolframalpha.com/input/?i=2*W%28-1%2F2%29

Answer 40

Ugghh.. I haven't a clue on how all that happened @.@
Don't bother explaining, I've already wasted your time xD

Answer 41

thanks guys, I was thoroughly curious!