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Question-174287




Question Number 174287 by Engr_Jidda last updated on 28/Jul/22
Answered by haf last updated on 28/Jul/22
use  ln function
$${use}\:\:{ln}\:{function} \\ $$
Commented by Engr_Jidda last updated on 29/Jul/22
pls help me out
$${pls}\:{help}\:{me}\:{out} \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jul/22
2^(x+4) +4^(x−2) =8  2^(x+4) +2^(2x−4) =8  2^4 ∙2^x +(2^x )^2 /2^4 =8  Let 2^x =y  16y+(y^2 /(16))=8  y^2 +256y−128=0  y=((−256±(√(256^2 −4(1)(−128))) )/2)     =−128±8(√(258))  2^x =128+(√(258))  xlog 2=log(−128±8(√(258)) )  x=((log(−128±8(√(258)) ))/(log 2))
$$\mathrm{2}^{{x}+\mathrm{4}} +\mathrm{4}^{{x}−\mathrm{2}} =\mathrm{8} \\ $$$$\mathrm{2}^{{x}+\mathrm{4}} +\mathrm{2}^{\mathrm{2}{x}−\mathrm{4}} =\mathrm{8} \\ $$$$\mathrm{2}^{\mathrm{4}} \centerdot\mathrm{2}^{{x}} +\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} /\mathrm{2}^{\mathrm{4}} =\mathrm{8} \\ $$$${Let}\:\mathrm{2}^{{x}} ={y} \\ $$$$\mathrm{16}{y}+\frac{{y}^{\mathrm{2}} }{\mathrm{16}}=\mathrm{8} \\ $$$${y}^{\mathrm{2}} +\mathrm{256}{y}−\mathrm{128}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{256}\pm\sqrt{\mathrm{256}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{128}\right)}\:}{\mathrm{2}} \\ $$$$\:\:\:=−\mathrm{128}\pm\mathrm{8}\sqrt{\mathrm{258}} \\ $$$$\mathrm{2}^{{x}} =\mathrm{128}+\sqrt{\mathrm{258}} \\ $$$${x}\mathrm{log}\:\mathrm{2}=\mathrm{log}\left(−\mathrm{128}\pm\mathrm{8}\sqrt{\mathrm{258}}\:\right) \\ $$$${x}=\frac{\mathrm{log}\left(−\mathrm{128}\pm\mathrm{8}\sqrt{\mathrm{258}}\:\right)}{\mathrm{log}\:\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 31/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 31/Jul/22
Thanks miss!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{miss}}! \\ $$

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