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Question Number 5568 by Rasheed Soomro last updated on 20/May/16

Solve without using calculator 4^x −3^(x+(1/2)) =3^(x−(1/2)) −2^(2x−1)

$$\mathrm{Solve}\:\mathrm{without}\:\mathrm{using}\:\mathrm{calculator} \\ $$$$\mathrm{4}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{3}^{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{2}^{\mathrm{2x}−\mathrm{1}} \\ $$

Answered by Yozzii last updated on 20/May/16

4^x +2^(2x−1) =3^x 3^(−0.5) +3^x ×3^(0.5) 4^x +4^x ×0.5=3^x ((1/(√3))+(√3)) ((4/3))^x =((√3)+(1/(√3)))×(2/3) x=((ln((8/3(√3))))/(ln(4/3))) x=((ln((4/3))+ln(2/(√3)))/(ln(4/3))) x=1+((0.5ln(4/3))/(ln(4/3))) x=1+0.5 x=1.5

$$\mathrm{4}^{{x}} +\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} =\mathrm{3}^{{x}} \mathrm{3}^{−\mathrm{0}.\mathrm{5}} +\mathrm{3}^{{x}} ×\mathrm{3}^{\mathrm{0}.\mathrm{5}} \\ $$$$\mathrm{4}^{{x}} +\mathrm{4}^{{x}} ×\mathrm{0}.\mathrm{5}=\mathrm{3}^{{x}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+\sqrt{\mathrm{3}}\right) \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)×\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}=\frac{{ln}\left(\left(\mathrm{8}/\mathrm{3}\sqrt{\mathrm{3}}\right)\right)}{{ln}\left(\mathrm{4}/\mathrm{3}\right)} \\ $$$${x}=\frac{{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+{ln}\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}}{{ln}\left(\mathrm{4}/\mathrm{3}\right)} \\ $$$${x}=\mathrm{1}+\frac{\mathrm{0}.\mathrm{5}{ln}\frac{\mathrm{4}}{\mathrm{3}}}{{ln}\left(\mathrm{4}/\mathrm{3}\right)} \\ $$$${x}=\mathrm{1}+\mathrm{0}.\mathrm{5} \\ $$$${x}=\mathrm{1}.\mathrm{5} \\ $$

Commented by Rasheed Soomro last updated on 21/May/16

Solution without logarthm. From above ((4/3))^x =((√3)+(1/(√3)))×(2/3) ((4/3))^x =((√3)+(1/(√3))×((√3)/(√3)))×(2/3) ((4/3))^x =((√3)+((√3)/3))×(2/3) ((4/3))^x =(1+(1/3))×((2(√3))/3) ((4/3))^x =(4/3)×((2(√3))/3) ((4/3))^(x−1) =(2/3^(1−1/2) )=(4^(1/2) /3^(1/2) )=((4/3))^(1/2) x−1=(1/2) x=1+(1/2)=(3/2)

$$\mathrm{Solution}\:\mathrm{without}\:\mathrm{logarthm}. \\ $$$$\mathrm{From}\:\mathrm{above} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)×\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\right)×\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\left(\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)×\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)×\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{x}} =\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{1}−\mathrm{1}/\mathrm{2}} }=\frac{\mathrm{4}^{\mathrm{1}/\mathrm{2}} }{\mathrm{3}^{\mathrm{1}/\mathrm{2}} }=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\mathrm{x}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Yozzii last updated on 20/May/16

Made an error. Sorry(√)

$${Made}\:{an}\:{error}.\:{Sorry}\sqrt{} \\ $$

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