sasogeek
  • sasogeek
\(\huge 3^x - 2^{y+2}=49 \) \(\huge 2^y - 3^{x-2}+1=0 \) solve the system of equations.
Mathematics
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sasogeek
  • sasogeek
\(\huge 3^x - 2^{y+2}=49 \) \(\huge 2^y - 3^{x-2}+1=0 \) solve the system of equations.
Mathematics
jamiebookeater
  • jamiebookeater
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sasogeek
  • sasogeek
i tried using logs but i got to a dead end... :/
asnaseer
  • asnaseer
try using a subtle substitution
sasogeek
  • sasogeek
how?

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asnaseer
  • asnaseer
\[a=3^x\]\[b=2^y\]
asnaseer
  • asnaseer
remember that \(2^{y+2}=2^2\times2^y=4\times2^y\)
sasogeek
  • sasogeek
ohhh, ok :) yeah...
asnaseer
  • asnaseer
:)
TuringTest
  • TuringTest
This is why I love this site, I never would have known what to do there...
sasogeek
  • sasogeek
i know right, i'm helping a friend out but i'm stuck myself so thanks to you guys ;)
asnaseer
  • asnaseer
I agree - I have learnt SO MUCH from this site - it is truly amazing.
anonymous
  • anonymous
where's the button for "round of applause"?
sasogeek
  • sasogeek
lol
asnaseer
  • asnaseer
he he :D
sasogeek
  • sasogeek
thanks guys, we did it! xD \(\large x \approx 4.069 \) and \(\large y \approx 3.263\) can someone confirm this please... :)
asnaseer
  • asnaseer
hmmm - I actually get integer solutions
asnaseer
  • asnaseer
Using the substitutions I listed above, I get:\[a-4b=49\tag{1}\]\[b-\frac{a}{9}+1=0\tag{2}\]Multiplying (2) by 9 and moving the constant to the right-hand-side, we get:\[9b-a=-9\tag{3}\]Adding (1) and (3) gives:\[5b=40\implies b=8\]Substituting this into (1) gives:\[a=81\]Therefore:\[2^y=8\implies y=3\]\[3^x=81\implies x=4\]

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