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solve algebraically: 2^(2x) + 2^X = 20?
I know the answer is x=2. I can solve it by trial and error and graphically. How is it solved algebraically with logs ?
4 Answers
- 1 decade agoFavorite Answer
take k=2^x. then 2^(2x)=k^2,then the eqn is k+k^2=20;
(k-4)(k+5)=0;
k=4;k=-5;
2^x=4,
x=2
- 1 decade ago
let y = 2^x
y^2 +y - 20 = 0
(y-4)(y+5) = 0
y= 4 or y = -5
2^x = 4 or 2^x = -5
x = 2 or
x = log(base 2)[-5] (reject this answer since it didn't give a real value)
- Jerome JLv 71 decade ago
2^(2x) + 2^x = 20
2^(2x) + 2^x - 20 = 0
(2^x + 5)(2^x - 4) = 0
(2^x + 5) = 0
2^x = -5 No. solution (no power of 2 could make -5)
(2^x - 4) = 0
2^x = 4
2^x = 2^2
x = 2
Or using logs
2^x = 4
ln 2^x = ln 4
x ln 2 = ln 4
x = ln 4/ln 2 = ln 2^2/ln 2 = 2ln 2/ln 2 = 2
- sahsjingLv 71 decade ago
(2^x)^2 + 2^x - 20 = (2^x + 5)(2^x - 4) = 0
Since 2^x + 5 > 0, we must have 2^x - 4 = 0 => 2^x = 2^2
Answer: x = 2
