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Question Number 198809 by dimentri last updated on 24/Oct/23

$$sumofroots$$$$\log {}_{3}x+\log {}_3(2,5)+\log {}_{x}9=3+\log {}_{x}5$$

Answered by a.lgnaoui last updated on 24/Oct/23

$$\frac{\mathrm{logx}}{\log 3}+\frac{\log (2,5)}{\log 3}+\frac{\log 9}{\mathrm{logx}}=3+\frac{\log 5}{\mathrm{logx}}$$$${\left(\mathrm{logx}\right)}^2+\mathrm{logx}.\log (2,5)+2{\left(\log 3\right)}^2$$$$=3\log 3\mathrm{logx}+\log 3.\log 5$$$$\mathrm{posons}\mathrm{logx}=\mathbf{z}$$$${\mathbf{z}}^2+\mathbf{z}[\mathbf{log}(2,5)-3\mathbf{log}3]+\mathbf{log}3(2\mathbf{log}3-\mathbf{log}5)=0$$$$$$$$△={[\mathbf{log}(2,5)-3\mathbf{log}3]}^2-4\mathbf{log}3(2\mathbf{log}3-\mathbf{log}5)$$$$\mathbf{log}(2,5)=\log \left(\frac{25}{10}\right)=\log 25-1=2\log 5-1$$$${\mathbf{z}}^2+\mathbf{z}(2\mathbf{log}5-3\mathbf{log}3-1)+\mathbf{log}3(2\mathbf{log}3-\mathbf{log}5)=0$$$$$$$$△={(2\log 5-3\log 3-1)}^2-8{\left(\log 3\right)}^2+$$$$4\log 3.\log 5$$$$=1+{\left(\log 3\right)}^2-16\mathbf{log}3\mathbf{log}5+6\mathbf{log}3$$$$+4{\left(\mathbf{log}5\right)}^2-4\mathbf{log}5$$$$={(2\mathbf{log}5-1)}^2+{(\mathbf{log}3-8\mathbf{log}5)}^2$$$$=+6\log 3-64{\left(\log 5\right)}^2$$$$$$$$=-6\mathbf{log}5-1)(10\mathbf{log}5-1)+$$$$$$$$△=6\mathbf{log}3-(10\mathbf{log}5-1)(6\mathbf{log}5+1)+$$$${(8\mathbf{log}5-\mathbf{log}3)}^2$$$$$$$$\mathbf{z}=\frac{(3\mathbf{log}3+1-2\mathbf{log}5)\pm \sqrt{6\mathbf{log}3-(6\mathbf{log}5+1)(10\mathbf{log}5-1)+{(8\mathbf{log}5-\mathbf{log}3)}^2}}{2}$$$$$$$$$$$$\mathbf{x}={\mathbf{e}}^{(3\mathbf{log}3-2\mathbf{log}5+1)/2}\times {\mathrm{e}}^{\pm \sqrt{6\log 3+{(8\log 5-3)}^2-6(\log 5-1)(6\mathrm{kog}5+1)}/2}$$$$$$$$\Rightarrow \mathbf{Sum}\mathrm{of}\mathbf{roots}\mathbf{is}$$$$\mathbf{r}={\mathbf{e}}^{\frac{3}{2}\mathbf{log}3-\mathbf{log}5+\frac{1}{2}}$$

Commented by mr W last updated on 24/Oct/23

$$wrong!$$$$x=e^{a\pm b}$$$$\Rightarrow \mathrm{\Sigma }x=e^a(e^b+e^{-b})\ne e^a$$

Commented by a.lgnaoui last updated on 24/Oct/23

$$\mathrm{yes}\mathrm{thanks}$$$$\mathrm{i}\mathrm{use}\mathrm{logaritme}\mathrm{base}10$$$$\mathrm{i}\mathrm{think}\mathrm{that}\mathrm{my}\mathrm{answer}\mathrm{wil}\mathrm{be}\mathrm{exact}$$$$\mathrm{if}\mathrm{the}\mathrm{we}\mathrm{change}\log \mathrm{by}\ln \mathrm{in}\mathrm{the}$$$$\mathrm{given}\mathrm{equation}.({\mathrm{c}}^{\prime }\mathrm{est}\mathrm{un}\mathrm{piege}!)$$$$$$

Commented by mr W last updated on 24/Oct/23

$$oui!$$

Answered by mr W last updated on 24/Oct/23

$$\log {}_{3}x+\log {}_3(2,5)-{\log }_3{3}^3=\log {}_{x}5-{\log }_{x}9$$$${\log }_3\frac{2.5x}{27}={\log }_x\frac{5}{9}$$$$\frac{\ln x+\ln \left(\frac{2.5}{27}\right)}{\log 3}=\frac{\ln \frac{5}{9}}{\ln x}$$$$\left(\ln x\right)(\ln x+\ln \frac{2.5}{27})=\left(\ln 3\right)\left(\ln \frac{5}{9}\right)$$$$lett=\ln x$$$$t^2+\left(\ln \frac{2.5}{27}\right)t-\left(\ln 3\right)\left(\ln \frac{5}{9}\right)=0$$$$t_{1,2}=\frac{1}{2}[-\ln \frac{2.5}{27}\pm \sqrt{{\left(\ln \frac{2.5}{27}\right)}^2+4\left(\ln 3\right)\left(\ln \frac{5}{9}\right)}]$$$$t_1+t_2=-\ln \frac{2.5}{27}=\ln \frac{54}{5}$$$$x_{1,2}=e^{t_{1,2}}$$$$\underset{―}{sumofroots:}$$$$x_1+x_2=e^{t_1}+e^{t_2}=....$$$$$$$$\underset{―}{productofroots:}$$$$x_1{x}_2=e^{t_1}{e}^{t_2}=e^{t_1+t_2}=e^{\ln \frac{54}{5}}=\frac{54}{5}$$

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