[แมวสามสี]

ข้อ 4 เปลี่ยนฐานเลยนะครับ

$\frac{log\frac{3}{x}}{log 3x}+(\frac{log x}{log 3})^2 = 1$

$\frac{log3-log x}{log3+log x}+\frac{(log x)^2}{(log 3)^2} = 1$

$(log3)^3 - (log3)^2 (log x) + (log3)(log x)^2 + (log x)^3 = (log3)^3 + (log3)^2 (log x) ; x\not= \frac{1}{3}$

จัดรูปใหม่ได้

$(logx)^3 - 2(logx)(log 3)^2 + (logx)^2 (log 3) = 0 $

$(logx)((logx)^2 - 2(log 3)^2 + (logx)(log 3)) = 0 $


$(logx)(logx + 2log 3)(logx - log 3) = 0 $