Solve this :
![2 {}^{x} + \frac{2}{2 {}^{x} } = 3 2 {}^{x} + \frac{2}{2 {}^{x} } = 3](https://tex.z-dn.net/?f=2+%7B%7D%5E%7Bx%7D++%2B++%5Cfrac%7B2%7D%7B2+%7B%7D%5E%7Bx%7D+%7D++%3D+3)
I will mark as brainliest answer.
Answers
Answered by
2
Final Answer : x = 0,1
Steps :
1) Let
![{2}^{x} = t {2}^{x} = t](https://tex.z-dn.net/?f=+%7B2%7D%5E%7Bx%7D+%3D+t)
then we get,
![t + \frac{2}{t} = 3 \\ = > {t}^{2} - 3t + 2 = 0 \\ = > (t - 2)(t - 1) = 0 \\ = > t = 1 \: or \: t \: = 2 t + \frac{2}{t} = 3 \\ = > {t}^{2} - 3t + 2 = 0 \\ = > (t - 2)(t - 1) = 0 \\ = > t = 1 \: or \: t \: = 2](https://tex.z-dn.net/?f=t+%2B+%5Cfrac%7B2%7D%7Bt%7D+%3D+3+%5C%5C+%3D+%26gt%3B+%7Bt%7D%5E%7B2%7D+-+3t+%2B+2+%3D+0+%5C%5C+%3D+%26gt%3B+%28t+-+2%29%28t+-+1%29+%3D+0+%5C%5C+%3D+%26gt%3B+t+%3D+1+%5C%3A+or+%5C%3A+t+%5C%3A+%3D+2)
Now,
![{2}^{x} = 1 = > x = 0 \\ {2}^{x} = 2 = > x = 1 {2}^{x} = 1 = > x = 0 \\ {2}^{x} = 2 = > x = 1](https://tex.z-dn.net/?f=+%7B2%7D%5E%7Bx%7D+%3D+1+%3D+%26gt%3B+x+%3D+0+%5C%5C+%7B2%7D%5E%7Bx%7D+%3D+2+%3D+%26gt%3B+x+%3D+1+)
Therefore, either x = 0 or x = 1 .
Steps :
1) Let
then we get,
Now,
Therefore, either x = 0 or x = 1 .
Similar questions
English,
9 months ago
Math,
9 months ago
Math,
1 year ago
Social Sciences,
1 year ago
English,
1 year ago
Social Sciences,
1 year ago