Question

4^1/x + 6^1/x - 9^1/x = 0

Помогите решить срочно

Answer 1

$4^{1/x}+6^{1/x}-9^{1/x}=0; ,; ; xne 0\\(2^{1/x})^2+2^{1/x}cdot 3^{1/x}-(3^{1/x})^2=0; |:(3^{1/x})^2ne 0\\t=(frac{2}{3})^{1/x}>0; ,; ; t^2+t-1=0; ,; ; (t+0,5)^2-1,25=0; ,$

$(t+0,5-sqrt{1,25})(t+0,5+sqrt{1,25})=0\\t_1=sqrt{1,25}-0,5>0; ; ,; ; t_2=-0,5-sqrt{1,25}<0; ; ne; podxodit$

$(frac{2}{3})^{1/x}=sqrt{frac{125}{100}}-frac{1}{2}; ,; ; (frac{2}{3})^{1/x}=frac{5sqrt5}{10}-frac{1}{2}=frac{sqrt5-1}{2}$

$frac{1}{x}=log_{2/3}frac{sqrt5-1}{2}; ,; ; x=frac{1}{log_{2/3}frac{sqrt5-1}{2}}$

$x=frac{log_2frac{2}{3}}{log_2(sqrt{5}-1)-log_22}=frac{1-log_23}{log_2(sqrt5-1)-1}$