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Question Number 40544 by mondodotto@gmail.com last updated on 23/Jul/18

solve for x  3^((2x−1)) −4(3^x )+1=0

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\mathrm{3}^{\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)} −\mathrm{4}\left(\mathrm{3}^{\boldsymbol{{x}}} \right)+\mathrm{1}=\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

3^(2x) .3^(−1) −4.3^x +1=0  a=3^x   a^2 ×(1/3)−4a+1=0  a^2 −12a+3=0  a=((12±(√(144−12)) )/2)=((12±(√(132)) )/2)  a=((12±2(√(33)) )/2)=6±(√(33))  3^x =6±(√(33))   x=log_3 (6±(√(33)) )

$$\mathrm{3}^{\mathrm{2}{x}} .\mathrm{3}^{−\mathrm{1}} −\mathrm{4}.\mathrm{3}^{{x}} +\mathrm{1}=\mathrm{0} \\ $$$${a}=\mathrm{3}^{{x}} \\ $$$${a}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{4}{a}+\mathrm{1}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{12}{a}+\mathrm{3}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{12}\pm\sqrt{\mathrm{144}−\mathrm{12}}\:}{\mathrm{2}}=\frac{\mathrm{12}\pm\sqrt{\mathrm{132}}\:}{\mathrm{2}} \\ $$$${a}=\frac{\mathrm{12}\pm\mathrm{2}\sqrt{\mathrm{33}}\:}{\mathrm{2}}=\mathrm{6}\pm\sqrt{\mathrm{33}} \\ $$$$\mathrm{3}^{{x}} =\mathrm{6}\pm\sqrt{\mathrm{33}}\: \\ $$$${x}={log}_{\mathrm{3}} \left(\mathrm{6}\pm\sqrt{\mathrm{33}}\:\right) \\ $$

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