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Question Number 50080 by F_Nongue last updated on 13/Dec/18
a) if f(x)=log(x+2), solve the equation: 2^(f(x−2)) ×2^(f(2x+2)) =4^(logf(x))
$$\left.{a}\right)\:{if}\:{f}\left({x}\right)={log}\left({x}+\mathrm{2}\right),\:{solve}\:{the}\:{equation}: \\ $$$$\mathrm{2}^{{f}\left({x}−\mathrm{2}\right)} ×\mathrm{2}^{{f}\left(\mathrm{2}{x}+\mathrm{2}\right)} =\mathrm{4}^{{logf}\left({x}\right)} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18
f(x−2)=log(x−2+2)=logx f(2x+2)=log(2x+2+2)=log(2x+4) logf(x)=log{log(x+2)+2} 2^(logx) ×2^(log(2x+4)) =2^(2log{log(x+2)+2}) logx+log(2x+4)=2log{log(x+2)+2} log{x(2x+4)}=log{log(x+2)+2}^2 2x^2 +4={log(x+2)+2}^2 2x^2 +4=4+4log(x+2)+{log(x+2)}^2 2x^2 =4log(x+2)+{log(x+2)}^2 k^2 +4k−2x^2 =0 k=log(x+2) k^2 +4k+4=2x^2 +4 (k+2)^2 =2(x^2 +2) k+2=(√(2(x^2 +2))) log(x+2)+2=(√(2(x^2 +2)))
$${f}\left({x}−\mathrm{2}\right)={log}\left({x}−\mathrm{2}+\mathrm{2}\right)={logx} \\ $$$${f}\left(\mathrm{2}{x}+\mathrm{2}\right)={log}\left(\mathrm{2}{x}+\mathrm{2}+\mathrm{2}\right)={log}\left(\mathrm{2}{x}+\mathrm{4}\right) \\ $$$${logf}\left({x}\right)={log}\left\{{log}\left({x}+\mathrm{2}\right)+\mathrm{2}\right\} \\ $$$$\mathrm{2}^{{logx}} ×\mathrm{2}^{{log}\left(\mathrm{2}{x}+\mathrm{4}\right)} =\mathrm{2}^{\mathrm{2}{log}\left\{{log}\left({x}+\mathrm{2}\right)+\mathrm{2}\right\}} \\ $$$${logx}+{log}\left(\mathrm{2}{x}+\mathrm{4}\right)=\mathrm{2}{log}\left\{{log}\left({x}+\mathrm{2}\right)+\mathrm{2}\right\} \\ $$$${log}\left\{{x}\left(\mathrm{2}{x}+\mathrm{4}\right)\right\}={log}\left\{{log}\left({x}+\mathrm{2}\right)+\mathrm{2}\right\}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}=\left\{{log}\left({x}+\mathrm{2}\right)+\mathrm{2}\right\}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{4}+\mathrm{4}{log}\left({x}+\mathrm{2}\right)+\left\{{log}\left({x}+\mathrm{2}\right)\right\}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} =\mathrm{4}{log}\left({x}+\mathrm{2}\right)+\left\{{log}\left({x}+\mathrm{2}\right)\right\}^{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} +\mathrm{4}{k}−\mathrm{2}{x}^{\mathrm{2}} =\mathrm{0}\:\:\:{k}={log}\left({x}+\mathrm{2}\right) \\ $$$${k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{4}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4} \\ $$$$\left({k}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$${k}+\mathrm{2}=\sqrt{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}\: \\ $$$${log}\left({x}+\mathrm{2}\right)+\mathrm{2}=\sqrt{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}\: \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18
Commented by Pk1167156@gmail.com last updated on 14/Dec/18
nice sir

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