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sasogeek

  • 3 years ago

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

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  1. sasogeek
    • 3 years ago
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    i tried using logs but i got to a dead end... :/

  2. asnaseer
    • 3 years ago
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    try using a subtle substitution

  3. sasogeek
    • 3 years ago
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    how?

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

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

  6. sasogeek
    • 3 years ago
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    ohhh, ok :) yeah...

  7. asnaseer
    • 3 years ago
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    :)

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

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

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

  11. JakeV8
    • 3 years ago
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    where's the button for "round of applause"?

  12. sasogeek
    • 3 years ago
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    lol

  13. asnaseer
    • 3 years ago
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    he he :D

  14. sasogeek
    • 3 years ago
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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... :)

  15. asnaseer
    • 3 years ago
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    hmmm - I actually get integer solutions

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

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