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Question-160342




Question Number 160342 by alf123 last updated on 28/Nov/21
Answered by Rasheed.Sindhi last updated on 28/Nov/21
4^x =6; ((8^x +8^(−x) )/(4^x +4^(−x) ))=?  8=4^(3/2)   ((8^x +8^(−x) )/(4^x +4^(−x) ))=(((4^(3/2) )^x +(4^(3/2) )^(−x) )/(4^x +4^(−x) ))  =(((4^x )^(3/2) +(4^x )^(−3/2) )/(4^x +(4^x )^(−1) ))=((6^(3/2) +6^(−3/2) )/(6+6^(−1) ))  =((6(√6) +(6(√6) )^(−1) )/(6(1/6)))=((6(6(√6) +(6(√6) )^(−1) ))/(37))  =((36(√6)+((√6) )^(−1) )/(37))=((36(√6) +(1/( (√6))))/(37))  =((216+1)/(37(√6)))×((√6)/( (√6)))=((217(√6))/(222))
$$\mathrm{4}^{{x}} =\mathrm{6};\:\frac{\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} }{\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} }=? \\ $$$$\mathrm{8}=\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{8}^{{x}} +\mathrm{8}^{−{x}} }{\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} }=\frac{\left(\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{{x}} +\left(\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)^{−{x}} }{\mathrm{4}^{{x}} +\mathrm{4}^{−{x}} } \\ $$$$=\frac{\left(\mathrm{4}^{{x}} \right)^{\mathrm{3}/\mathrm{2}} +\left(\mathrm{4}^{{x}} \right)^{−\mathrm{3}/\mathrm{2}} }{\mathrm{4}^{{x}} +\left(\mathrm{4}^{{x}} \right)^{−\mathrm{1}} }=\frac{\mathrm{6}^{\mathrm{3}/\mathrm{2}} +\mathrm{6}^{−\mathrm{3}/\mathrm{2}} }{\mathrm{6}+\mathrm{6}^{−\mathrm{1}} } \\ $$$$=\frac{\mathrm{6}\sqrt{\mathrm{6}}\:+\left(\mathrm{6}\sqrt{\mathrm{6}}\:\right)^{−\mathrm{1}} }{\mathrm{6}\frac{\mathrm{1}}{\mathrm{6}}}=\frac{\mathrm{6}\left(\mathrm{6}\sqrt{\mathrm{6}}\:+\left(\mathrm{6}\sqrt{\mathrm{6}}\:\right)^{−\mathrm{1}} \right)}{\mathrm{37}} \\ $$$$=\frac{\mathrm{36}\sqrt{\mathrm{6}}+\left(\sqrt{\mathrm{6}}\:\right)^{−\mathrm{1}} }{\mathrm{37}}=\frac{\mathrm{36}\sqrt{\mathrm{6}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}}{\mathrm{37}} \\ $$$$=\frac{\mathrm{216}+\mathrm{1}}{\mathrm{37}\sqrt{\mathrm{6}}}×\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{6}}}=\frac{\mathrm{217}\sqrt{\mathrm{6}}}{\mathrm{222}} \\ $$
Commented by MJS_new last updated on 28/Nov/21
good!  ...=((217(√6))/(222))
$$\mathrm{good}! \\ $$$$…=\frac{\mathrm{217}\sqrt{\mathrm{6}}}{\mathrm{222}} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Nov/21
Thanks sir!  I′ve simplified now.
$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{sir}! \\ $$$${I}'{ve}\:{simplified}\:{now}. \\ $$

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