The number of solutions of equation `log_3 x log_4 x log_5 x= log_3 x log_4 x +log_5 x`

Explanation

Let `log_5x = t`, then, `x = 5^t`

`=>log_3x = log_3 5^t = tlog_3 5`

`=>log_4x = log_4 5^t = tlog_4 5`

So, our equation becomes,

`tlog_3 5*tlog_4 5*t = tlog_3 5*tlog_4 +t`

`=>t^3(log_3 5log_4 5) - t^2(log_3 5log_4 5) -t = 0`

`=>t(t^2(log_3 5log_4 5) - t(log_3 5log_4 5) -1) = 0`

`=>t = 0 or t^2(log_3 5log_4 5) - t(log_3 5log_4 5) -1 = 0`

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