2-2a-2-a-10-what-is-a-

Question Number 175849 by Linton last updated on 08/Sep/22

2^{2 a} + 2^{a} = 10

w h a t i s a ?

Answered by MJS_new last updated on 08/Sep/22

2^{2 a} + 2^{a} - 10 = 0

{(2^{a})}^{2} + 2^{a} - 10 = 0

2^{a} = - \frac{1}{2} \pm \sqrt{\frac{1}{4} + 10} = - \frac{1}{2} \pm \frac{\sqrt{41}}{2}

2^{a} > 0 \forall a \in R \Rightarrow 2^{a} = \frac{- 1 + \sqrt{41}}{2}

a = \frac{\ln \frac{- 1 + \sqrt{41}}{2}}{\ln 2} = \frac{\ln (- 1 + \sqrt{41})}{\ln 2} - 1

Answered by BaliramKumar last updated on 08/Sep/22

2^{2 a} + 2^{a} = 10

2^{a} (2^{a} + 1) - 10 = 0 p u t 2^{a} = x

x (x + 1) - 10 = 0

x^{2} + x - 10 = 0

x = \frac{- 1 \pm \sqrt{1 + 40}}{2} = \frac{- 1 + \sqrt{41}}{2} (+ ve)

2^{a} = (\frac{\sqrt{41} - 1}{2})

a \cdot l o g (2) = l o g (\frac{\sqrt{41} - 1}{2})

a = \frac{l o g (\frac{\sqrt{41} - 1}{2})}{l o g (2)}

a = \log_{2} (\frac{\sqrt{41} - 1}{2}) A n s w e r