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 $
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