Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

JonaskBest ResponseYou've already chosen the best response.0
a)\[3^x=2^xy+1\] b)\[x^y=y^{xy}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
3^x = 2^x y + 1 y =0, x=0 < this works ... since the lhs is power of 3
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
x=1, y=1 also works where you will get 3 = 3
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
so we just try out numbers
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
no ...we can't do that. diophantine equations have always been my weaknesses. I'm just fooling around.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
if 3^n 1 is of form divisible by 2^n, then we have solutions of it.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
actually i never knew that they are called that diophantine
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[3^x2^xy=1\] but we can try x=1,2,3,4 and see if they yield the y value to give 1
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
for these types of question, you should probably ask mukushla till then let me fool around.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
dw:1350060934777:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
dw:1350061042671:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
If, every C(n,i) is divisible by 2^(ni) then we have integer solutions. we can check this manually ... since for n>N, this will be invalid.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
for the first one x=1 is answer for x>1 conclude that x must be even from\[3^{x1}+3^{x2}+...+3+1=2^{x1}y\]and setting x=2k going back to original equation\[3^{2k}1=2^{2k}y\]\[(3^k1)(3^k+1)=2^{2k}y\]now since \(\gcd(3^k1,3^k+1)=2\) one of \(3^k1 \ \ , \ 3^k+1\) must be divisible by \(2^{2k1}\) and this is possible onle when \(k<3\)
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
for the second one note that\[(\frac{x}{y})^y=y^{x2y}\] so \(x\ge y\) and \(yx\) so set \[x=ky\]and find the limits for k
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
if u wanna check the solutions are for the first one (x,y)=(1,1),(2,2),(4,5) for the second one (x,y)=(1,1),(9,3),(8,2)
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
thanks man!! you always amaze me!!
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.