/sci/ - Science & Math

>>4433799

e^(5/x) + e^x = e^10

then take log of both sides

5/x + x = 10

you should be able to solve this

>>4433822

oops..

e^(5/x) + e^x = e^10 is actually

e^(5/x) + x = e^10

>>4433799

no freaken idea lol

>>4433822
>>4433824

What the fuck?

<span class="math">\displaystyle e^a*e^b = e^{a+b}[/spoiler]


<span class="math">\displaystyle \frac{5}{x}+log(x)=10[/spoiler]


<span class="math">\displaystyle e^{\frac{5}{x}+log(x)}=e^{10}[/spoiler]

<span class="math">\displaystyle e^{\frac{5}{x}}e^{log(x)}=e^{10}[/spoiler]

<span class="math">\displaystyle e^{\frac{5}{x}}x=e^{10}[/spoiler]

You would need to know how to use the Lambert-W function to proceed further.

>>4433877

> e^a * e^b=e^a+b

what does that have to do with anything

>>4433890

...

>e^(5/x) + e^x = e^10
>e^(5/x) + x = e^10

Find the error and know why it is linked to the exponential property I just listed and get a cookie.

>>4433890

He's just stating how you got there in-case you failed 10th grade maths

>>4433901

>>e^(5/x) + x = e^10

this is correct

there is no error here, thats how you exponentiate both sides

>>4433935
Don't fucking post math shit unless you know what the fuck you are talking about, and at lower levels, it should be clear when you don't know what you are talking about (now is one of those times). Exponentiation does not distribute over addition like you seem to think it does.

>>4433935
Underage b&.

>>4433935
Oh I see, so <span class="math">e^{5/x + logx}=e^{5/x} + e^{logx}[/spoiler]? That's not how math works mate.

>>4433935
Oh I see, so <span class="math">e^{5/x \ + \ logx}=e^{5/x} + e^{logx}[/spoiler]? That's not how math works mate.

>>4433947

show me a formal proof otherwise you are wrong,.

>>4433955

no one said those are equivalent, but when you exponentiate both sides you get the right hand side not the left hand side

>>4433956
>>4433956

no one said those are equivalent,
learn to read

Too much stupidity in this post over something as simple as exponential rules.

>>4433957
>show me why I am retarded and have forgotten exponential rules learned in high school
>OR YOU'RE WRONG!!!!11ONE

>>4433961
>"no one said they were equivalent"
>second and third post say they're equivalent
>"LEARN TO READ!!!!ONELEVEN!!"

>>4433975

so you can't prove it? k fair enough you dunno what you're talkin about


second post was corrected in the third

>e^(5/x) + e^x = e^10 is actually

>e^(5/x) + x = e^10

they aren't equivalent, the first one is wrong and "is actually e^(5/x) + x = e^10 is actually"

see if you learn to read? things would work out better
come back to me when you can prove your highschool mistakes are actual math

5/x+log(x)-10=0
Than do Newton's iteration infinite times.

ITT: derps who can't take two seconds to think through what they are saying.

1 + 1 = 2
1^(1 + 1) = 1^2
1 = 1

1 + 1 = 2
1^1 + 1^1 = 1^2
1 + 1 = 1
2 != 1

>>4433994

sure is highschooler in here

e^x doesn't behave like that

watch.

y = x
dy/dx = 1

y= e^x
dy/dx = e^x

dy/dx =/= 1.....

e^x is special.

>>4434041
If you aren't in high school, then I feel sorry for you. e^x is indeed special, but what you said makes absolutely no sense. It does not break any rules. What you just said makes absolutely no sense.

Just bumping this thread because I need it as an example of weak understanding of algebraic concepts for another thread. :)

You cannot isolate x since the equation involves a trascendent operator (log) summed to some algebraic form of x. You can only obtain approximated solutions using numerical methods.

This thread fills every once of me with fear and rage.

The equation is transcendental, there is no algebraic way to solve it, except through approximation. Alternatively, you could solve it by graphing it and finding the intersection of the graphs.

>>4435372

> This thread fills every once of me with fear and rage.
How many times, again?

>>4435403

Typos are serious buisness.

>>4435468
Like how I hit the i before the n in business.