Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

2^(2a) +2^a = 10  what is a?

$$\mathrm{2}^{\mathrm{2}{a}} +\mathrm{2}^{{a}} =\:\mathrm{10} \\ $$$${what}\:{is}\:{a}? \\ $$

Answered by MJS_new last updated on 08/Sep/22

2^(2a) +2^a −10=0  (2^a )^2 +2^a −10=0  2^a =−(1/2)±(√((1/4)+10))=−(1/2)±((√(41))/2)  2^a >0∀a∈R ⇒ 2^a =((−1+(√(41)))/2)  a=((ln ((−1+(√(41)))/2))/(ln 2))=((ln (−1+(√(41))))/(ln 2))−1

$$\mathrm{2}^{\mathrm{2}{a}} +\mathrm{2}^{{a}} −\mathrm{10}=\mathrm{0} \\ $$$$\left(\mathrm{2}^{{a}} \right)^{\mathrm{2}} +\mathrm{2}^{{a}} −\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{2}^{{a}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{10}}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$$$\mathrm{2}^{{a}} >\mathrm{0}\forall{a}\in\mathbb{R}\:\Rightarrow\:\mathrm{2}^{{a}} =\frac{−\mathrm{1}+\sqrt{\mathrm{41}}}{\mathrm{2}} \\ $$$${a}=\frac{\mathrm{ln}\:\frac{−\mathrm{1}+\sqrt{\mathrm{41}}}{\mathrm{2}}}{\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{ln}\:\left(−\mathrm{1}+\sqrt{\mathrm{41}}\right)}{\mathrm{ln}\:\mathrm{2}}−\mathrm{1} \\ $$

Answered by BaliramKumar last updated on 08/Sep/22

  2^(2a) + 2^a  = 10  2^a (2^a +1) −10=0     put     2^a  = x  x(x+1)−10 = 0  x^2 +x−10=0  x = ((−1±(√(1+40)))/2) = ((−1+(√(41)))/2) (+ve)  2^a  = ((((√(41))−1)/2))  a∙log(2) = log((((√(41))−1)/2))  a = ((log((((√(41))−1)/2)))/(log(2)))  a = log_2 ((((√(41))−1)/2)) Answer

$$ \\ $$$$\mathrm{2}^{\mathrm{2}{a}} +\:\mathrm{2}^{{a}} \:=\:\mathrm{10} \\ $$$$\mathrm{2}^{{a}} \left(\mathrm{2}^{{a}} +\mathrm{1}\right)\:−\mathrm{10}=\mathrm{0}\:\:\:\:\:{put}\:\:\:\:\:\mathrm{2}^{{a}} \:=\:{x} \\ $$$${x}\left({x}+\mathrm{1}\right)−\mathrm{10}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{10}=\mathrm{0} \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{40}}}{\mathrm{2}}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{41}}}{\mathrm{2}}\:\left(+\mathrm{ve}\right) \\ $$$$\mathrm{2}^{{a}} \:=\:\left(\frac{\sqrt{\mathrm{41}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${a}\centerdot{log}\left(\mathrm{2}\right)\:=\:{log}\left(\frac{\sqrt{\mathrm{41}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${a}\:=\:\frac{{log}\left(\frac{\sqrt{\mathrm{41}}−\mathrm{1}}{\mathrm{2}}\right)}{{log}\left(\mathrm{2}\right)} \\ $$$${a}\:=\:\mathrm{log}_{\mathrm{2}} \left(\frac{\sqrt{\mathrm{41}}−\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{A}{nswer} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com