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Question Number 146386 by mathdanisur last updated on 13/Jul/21

if   (8/(3^x  + 2^(−x) )) = ((27)/(3^(−x)  + 2^x ))  find   2x+1=?

$${if}\:\:\:\frac{\mathrm{8}}{\mathrm{3}^{\boldsymbol{{x}}} \:+\:\mathrm{2}^{−\boldsymbol{{x}}} }\:=\:\frac{\mathrm{27}}{\mathrm{3}^{−\boldsymbol{{x}}} \:+\:\mathrm{2}^{\boldsymbol{{x}}} } \\ $$$${find}\:\:\:\mathrm{2}{x}+\mathrm{1}=? \\ $$

Answered by iloveisrael last updated on 13/Jul/21

 If (8/(3^x +2^(−x) )) = ((27)/(3^(−x) +2^x ))    then the value of 2x+1 = __   ((8.2^x )/(6^x +1)) = ((27.3^x )/(6^x +1))   ⇒ 2^(x+3)  = 3^(x+3)  , x=−3  then 2x+1=−5

$$\:\mathrm{If}\:\frac{\mathrm{8}}{\mathrm{3}^{\mathrm{x}} +\mathrm{2}^{−\mathrm{x}} }\:=\:\frac{\mathrm{27}}{\mathrm{3}^{−\mathrm{x}} +\mathrm{2}^{\mathrm{x}} }\: \\ $$$$\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{2x}+\mathrm{1}\:=\:\_\_ \\ $$$$\:\frac{\mathrm{8}.\mathrm{2}^{\mathrm{x}} }{\mathrm{6}^{\mathrm{x}} +\mathrm{1}}\:=\:\frac{\mathrm{27}.\mathrm{3}^{\mathrm{x}} }{\mathrm{6}^{\mathrm{x}} +\mathrm{1}}\: \\ $$$$\Rightarrow\:\mathrm{2}^{\mathrm{x}+\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{x}+\mathrm{3}} \:,\:\mathrm{x}=−\mathrm{3} \\ $$$$\mathrm{then}\:\mathrm{2x}+\mathrm{1}=−\mathrm{5}\: \\ $$

Commented by mathdanisur last updated on 13/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

Answered by Rasheed.Sindhi last updated on 13/Jul/21

(8/(3^x  + 2^(−x) ))=((27)/(3^(−x)  + 2^x ))  (8/(3^x +(1/2^x )))=((27)/(2^x +(1/3^x )))  (8/((6^x +1)/2^x ))=((27)/((6^x +1)/3^x ))  8((2^x /(6^x +1)))=27((3^x /(6^x +1)))  (2^x /(6^x +1)).((6^x +1)/3^x )=((27)/8)  ((2/3))^x =((3/2))^3   ((2/3))^x =((2/3))^(−3)   x=−3  2x+1=2(−3)+1=−5

$$\frac{\mathrm{8}}{\mathrm{3}^{\boldsymbol{{x}}} \:+\:\mathrm{2}^{−\boldsymbol{{x}}} }=\frac{\mathrm{27}}{\mathrm{3}^{−\boldsymbol{{x}}} \:+\:\mathrm{2}^{\boldsymbol{{x}}} } \\ $$$$\frac{\mathrm{8}}{\mathrm{3}^{{x}} +\frac{\mathrm{1}}{\mathrm{2}^{{x}} }}=\frac{\mathrm{27}}{\mathrm{2}^{{x}} +\frac{\mathrm{1}}{\mathrm{3}^{{x}} }} \\ $$$$\frac{\mathrm{8}}{\frac{\mathrm{6}^{{x}} +\mathrm{1}}{\mathrm{2}^{{x}} }}=\frac{\mathrm{27}}{\frac{\mathrm{6}^{{x}} +\mathrm{1}}{\mathrm{3}^{{x}} }} \\ $$$$\mathrm{8}\left(\frac{\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} +\mathrm{1}}\right)=\mathrm{27}\left(\frac{\mathrm{3}^{{x}} }{\mathrm{6}^{{x}} +\mathrm{1}}\right) \\ $$$$\frac{\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} +\mathrm{1}}.\frac{\mathrm{6}^{{x}} +\mathrm{1}}{\mathrm{3}^{{x}} }=\frac{\mathrm{27}}{\mathrm{8}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{−\mathrm{3}} \\ $$$${x}=−\mathrm{3} \\ $$$$\mathrm{2}{x}+\mathrm{1}=\mathrm{2}\left(−\mathrm{3}\right)+\mathrm{1}=−\mathrm{5} \\ $$

Commented by mathdanisur last updated on 13/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

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