Menu Close

Question-205130




Question Number 205130 by Ari last updated on 09/Mar/24
Commented by Ari last updated on 09/Mar/24
x=?
$${x}=? \\ $$
Answered by A5T last updated on 09/Mar/24
3^x 2^x =3^(2x) −2^(2x)  ⇒((3^x 2^x )/2^(2x) )=((3/2))^(2x) −1  ⇒p^2 −p−1=0 where p=((3/2))^x   ⇒p=((1+_− (√5))/2)=((3/2))^x ⇒x=log_(3/2) (((1+(√5))/2))  x=((log(1+(√5))−log2)/(log3−log2))
$$\mathrm{3}^{{x}} \mathrm{2}^{{x}} =\mathrm{3}^{\mathrm{2}{x}} −\mathrm{2}^{\mathrm{2}{x}} \:\Rightarrow\frac{\mathrm{3}^{{x}} \mathrm{2}^{{x}} }{\mathrm{2}^{\mathrm{2}{x}} }=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}{x}} −\mathrm{1} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −{p}−\mathrm{1}=\mathrm{0}\:{where}\:{p}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \\ $$$$\Rightarrow{p}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{5}}}{\mathrm{2}}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \Rightarrow{x}={log}_{\frac{\mathrm{3}}{\mathrm{2}}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${x}=\frac{{log}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)−{log}\mathrm{2}}{{log}\mathrm{3}−{log}\mathrm{2}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *