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sasogeek
 3 years ago
\(\huge 3^x  2^{y+2}=49 \)
\(\huge 2^y  3^{x2}+1=0 \)
solve the system of equations.
sasogeek
 3 years ago
\(\huge 3^x  2^{y+2}=49 \) \(\huge 2^y  3^{x2}+1=0 \) solve the system of equations.

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sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0i tried using logs but i got to a dead end... :/

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4try using a subtle substitution

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4remember that \(2^{y+2}=2^2\times2^y=4\times2^y\)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0This is why I love this site, I never would have known what to do there...

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0i know right, i'm helping a friend out but i'm stuck myself so thanks to you guys ;)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4I agree  I have learnt SO MUCH from this site  it is truly amazing.

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0where's the button for "round of applause"?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0thanks guys, we did it! xD \(\large x \approx 4.069 \) and \(\large y \approx 3.263\) can someone confirm this please... :)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4hmmm  I actually get integer solutions

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4Using the substitutions I listed above, I get:\[a4b=49\tag{1}\]\[b\frac{a}{9}+1=0\tag{2}\]Multiplying (2) by 9 and moving the constant to the righthandside, we get:\[9ba=9\tag{3}\]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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