Question Number 190820 by Rupesh123 last updated on 12/Apr/23
Answered by Rasheed.Sindhi last updated on 12/Apr/23
![((3^(3x) −2^(3x) )/(2^x ∙3^(2x) −3^x ∙2^(2x) ))=(7/2) (((3^x )^3 −(2^x )^3 )/(2^x ∙3^x (3^x −2^x )))=(7/2) (((3^x −2^x )(3^(2x) +3^x 2^x +2^(2x) ))/(2^x ∙3^x (3^x −2^x )))=(7/2) (3^(2x) /(2^x ∙3^x ))+((2^x ∙3^x )/(2^x ∙3^x ))+(2^(2x) /(2^x ∙3^x ))=(7/2) ((3/2))^x +((2/3))^x =(7/2)−1=(5/2) y+(1/y)=(5/2) [let ((3/2))^x =y] 2y^2 −5y+2=0 y=((3/2))^x =((5±(√(25−16)) )/4)=((5±3)/4)=2,(1/2) log((3/2))^x =log 2 , log ((1/2)) x log((3/2))=log 2 , log ((1/2)) x=((log 2)/(log 3−log 2)) , ((log 1−log 2)/(log 3−log 2)) x=((log_2 2)/(log_2 3−log_2 2)) , ((log_2 1−log_2 2)/(log_2 3−log_2 2)) x=(1/(log_2 3−1)) , ((0−1)/(log_2 3−1)) x=±(1/(log_2 3−1))](https://www.tinkutara.com/question/Q190825.png)
$$\frac{\mathrm{3}^{\mathrm{3}{x}} −\mathrm{2}^{\mathrm{3}{x}} }{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{\mathrm{2}{x}} −\mathrm{3}^{{x}} \centerdot\mathrm{2}^{\mathrm{2}{x}} }=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\frac{\left(\mathrm{3}^{{x}} \right)^{\mathrm{3}} −\left(\mathrm{2}^{{x}} \right)^{\mathrm{3}} }{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} \left(\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \right)}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\frac{\cancel{\left(\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \right)}\left(\mathrm{3}^{\mathrm{2}{x}} +\mathrm{3}^{{x}} \mathrm{2}^{{x}} +\mathrm{2}^{\mathrm{2}{x}} \right)}{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} \cancel{\left(\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \right)}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}^{\mathrm{2}{x}} }{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} }+\frac{\cancel{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} }}{\cancel{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} }}+\frac{\mathrm{2}^{\mathrm{2}{x}} }{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} }=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${y}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{5}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left[{let}\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} ={y}\right] \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\mathrm{5}{y}+\mathrm{2}=\mathrm{0} \\ $$$${y}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} =\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}\:}{\mathrm{4}}=\frac{\mathrm{5}\pm\mathrm{3}}{\mathrm{4}}=\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{log}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} =\mathrm{log}\:\mathrm{2}\:,\:\mathrm{log}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${x}\:\mathrm{log}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{log}\:\mathrm{2}\:,\:\mathrm{log}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${x}=\frac{\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{3}−\mathrm{log}\:\mathrm{2}}\:\:,\:\:\frac{\mathrm{log}\:\mathrm{1}−\mathrm{log}\:\mathrm{2}}{\mathrm{log}\:\mathrm{3}−\mathrm{log}\:\mathrm{2}} \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \mathrm{2}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}−\mathrm{log}_{\mathrm{2}} \mathrm{2}}\:\:,\:\:\frac{\mathrm{log}_{\mathrm{2}} \mathrm{1}−\mathrm{log}_{\mathrm{2}} \mathrm{2}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}−\mathrm{log}_{\mathrm{2}} \mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}}\:\:,\:\:\frac{\mathrm{0}−\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}} \\ $$$${x}=\pm\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{3}−\mathrm{1}}\: \\ $$
Commented by Rupesh123 last updated on 12/Apr/23
Excellent, sir!
Commented by Spillover last updated on 12/Apr/23
$$\mathrm{good} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Apr/23
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}} \\ $$