若logx/2*log3/x=logx/6有兩個相異實根,則此兩根之積為?

log(x/2) * log(3/x) = log(x/6) ...... [1]
(logx - log2) * (log3 - logx) = logx - log6
log3 logx - (logx)² - log2 log3 + log2 logx - logx + log6 = 0
- (logx)² + (log3 + log2 - 1) logx - (log2 log3 + log6) = 0
(logx)² - (log3 + log2 - 1) logx + (log2 log3 + log6) = 0

設 u = logx，
u² - (log3 + log2 - 1) u + (log2 log3 + log6) = 0 ...... [2]
則logα 及 logβ 為方程式 [2] 的兩根。

方程式 [2] 的兩根之和：
logα + logβ = log3 + log2 - 1
log(αβ) = log3 + log2 - log10
log(αβ) = log(3*2/10)
log(αβ) = log(3/5)
αβ = 3/5

原方程式的兩根之積 αβ = 3/5

log(x/2) * log(3/x) = log(x/6)
(log x - log 2)(log 3 - log x) = log x - log 6
(log x)(log 3)-(log x)²-(log 3)(log 2)+(log x)(log 2)=log x -log 6
-(log x)²+(log x)(log 2)+(log x)(log 3)-log x+log 6-(log 3)(log 2)=0
log x=-0.91518034 or log x=0.69333159
x=0.121568108 or x=4.935504937
兩根之積為=0.121568108*4.935504937=0.6=3/5