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sasogeek Group Title

\(\huge 3^x - 2^{y+2}=49 \) \(\huge 2^y - 3^{x-2}+1=0 \) solve the system of equations.

  • 2 years ago
  • 2 years ago

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  1. sasogeek Group Title
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    i tried using logs but i got to a dead end... :/

    • 2 years ago
  2. asnaseer Group Title
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    try using a subtle substitution

    • 2 years ago
  3. sasogeek Group Title
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    how?

    • 2 years ago
  4. asnaseer Group Title
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    \[a=3^x\]\[b=2^y\]

    • 2 years ago
  5. asnaseer Group Title
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    remember that \(2^{y+2}=2^2\times2^y=4\times2^y\)

    • 2 years ago
  6. sasogeek Group Title
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    ohhh, ok :) yeah...

    • 2 years ago
  7. asnaseer Group Title
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    :)

    • 2 years ago
  8. TuringTest Group Title
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    This is why I love this site, I never would have known what to do there...

    • 2 years ago
  9. sasogeek Group Title
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    i know right, i'm helping a friend out but i'm stuck myself so thanks to you guys ;)

    • 2 years ago
  10. asnaseer Group Title
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    I agree - I have learnt SO MUCH from this site - it is truly amazing.

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  11. JakeV8 Group Title
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    where's the button for "round of applause"?

    • 2 years ago
  12. sasogeek Group Title
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    lol

    • 2 years ago
  13. asnaseer Group Title
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    he he :D

    • 2 years ago
  14. sasogeek Group Title
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    thanks guys, we did it! xD \(\large x \approx 4.069 \) and \(\large y \approx 3.263\) can someone confirm this please... :)

    • 2 years ago
  15. asnaseer Group Title
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    hmmm - I actually get integer solutions

    • 2 years ago
  16. asnaseer Group Title
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    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:\[2^y=8\implies y=3\]\[3^x=81\implies x=4\]

    • 2 years ago
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is replying to Can someone tell me what button the professor is hitting...

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