2^x + 2^y = 12 x + y = 5 How do you solve this algebraically (instead of guess and check)?
frome second equation we have y=5-x plug this into first one and solve for x
i got 2^5 = 12 which is a false statement... 2^x + 2^y = 12 x + y = 5 y = 5 - x 2^x + 2^5 - x = 12 2^x + 5 - x = 12 2^5 = 12 2^5 = 32 32 does not equal 12...
dude what did u do ??
i substituted y = 5^x in the first equation, making that part of the equatoin 2^5 - x
after sub we have\[2^x+2^{5-x}=12\]
\[2^x+\frac{2^5}{2^x}=12\]let \[z=2^x\]\[z+\frac{32}{z}=12\]multiply by z\[z^2-12z+32=0\]this is a quadratic solve for z and ...
oh i got it before i was using a rule for simplifying exponents (n^y x n^z = n^y = z) incorrectly, i was using it with addition which doesn't work. Thank you!
no problem
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