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Question Number 124880 by WilliamsErinfolami last updated on 06/Dec/20

Solve for x x^(log_5 3x) =12x

$${Solve}\:{for}\:{x} \\ $$$${x}^{{log}_{\mathrm{5}} \mathrm{3}{x}} =\mathrm{12}{x} \\ $$

Answered by mr W last updated on 06/Dec/20

(log_5 3x)(log_5 x)=log_5 (12x) (log_5 3+log_5 x)(log_5 x)=log_5 12+log_5 x (log_5 3+t)t=log_5 12+t t^2 −(1−log_5 3)t−log_5 12=0 t=((1−log_5 3±(√((1−log_5 3)^2 +4log_5 12)))/2) log_5 x=((1−log_5 3±(√((1−log_5 3)^2 +4log_5 12)))/2) ⇒x=5^((1−log_5 3±(√((1−log_5 3)^2 +4log_5 12)))/2) =9.693765, 0.171932

$$\left(\mathrm{log}_{\mathrm{5}} \:\mathrm{3}{x}\right)\left(\mathrm{log}_{\mathrm{5}} \:{x}\right)=\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{12}{x}\right) \\ $$$$\left(\mathrm{log}_{\mathrm{5}} \:\mathrm{3}+\mathrm{log}_{\mathrm{5}} \:{x}\right)\left(\mathrm{log}_{\mathrm{5}} \:{x}\right)=\mathrm{log}_{\mathrm{5}} \:\mathrm{12}+\mathrm{log}_{\mathrm{5}} \:{x} \\ $$$$\left(\mathrm{log}_{\mathrm{5}} \:\mathrm{3}+{t}\right){t}=\mathrm{log}_{\mathrm{5}} \:\mathrm{12}+{t} \\ $$$${t}^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\right){t}−\mathrm{log}_{\mathrm{5}} \:\mathrm{12}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\pm\sqrt{\left(\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4log}_{\mathrm{5}} \:\mathrm{12}}}{\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{5}} \:{x}=\frac{\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\pm\sqrt{\left(\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4log}_{\mathrm{5}} \:\mathrm{12}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{5}^{\frac{\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\pm\sqrt{\left(\mathrm{1}−\mathrm{log}_{\mathrm{5}} \:\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4log}_{\mathrm{5}} \:\mathrm{12}}}{\mathrm{2}}} \\ $$$$=\mathrm{9}.\mathrm{693765},\:\mathrm{0}.\mathrm{171932} \\ $$

Answered by aleks041103 last updated on 06/Dec/20

log_x (12x)=log_5 (3x) 1+log_x (12)=log_5 3 + log_5 x 1+((log_5 12)/(log_5 x)) = log_5 3 + log_5 x t=log_5 x ⇒ x=5^t t+log_5 12 = t^2 +(log_5 3)t ⇒t^2 −(1−log_5 3)t−log_5 12 =0 t_(1,2) =((1−log_5 3±(√((1−log_5 3)^2 +4log_5 12)))/2) ⇒x_(1,2) =5^t_(1,2)

$${log}_{{x}} \left(\mathrm{12}{x}\right)={log}_{\mathrm{5}} \left(\mathrm{3}{x}\right) \\ $$$$\mathrm{1}+{log}_{{x}} \left(\mathrm{12}\right)={log}_{\mathrm{5}} \mathrm{3}\:+\:{log}_{\mathrm{5}} {x} \\ $$$$\mathrm{1}+\frac{{log}_{\mathrm{5}} \mathrm{12}}{{log}_{\mathrm{5}} {x}}\:=\:{log}_{\mathrm{5}} \mathrm{3}\:+\:{log}_{\mathrm{5}} {x} \\ $$$${t}={log}_{\mathrm{5}} {x}\:\Rightarrow\:{x}=\mathrm{5}^{{t}} \\ $$$${t}+{log}_{\mathrm{5}} \mathrm{12}\:=\:{t}^{\mathrm{2}} +\left({log}_{\mathrm{5}} \mathrm{3}\right){t} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\left(\mathrm{1}−{log}_{\mathrm{5}} \mathrm{3}\right){t}−{log}_{\mathrm{5}} \mathrm{12}\:=\mathrm{0} \\ $$$${t}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}−{log}_{\mathrm{5}} \mathrm{3}\pm\sqrt{\left(\mathrm{1}−{log}_{\mathrm{5}} \mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}{log}_{\mathrm{5}} \mathrm{12}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{1},\mathrm{2}} =\mathrm{5}^{{t}_{\mathrm{1},\mathrm{2}} } \\ $$

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