Help Please!!!

Each of these logs can be written into log form (that is, log₁₀ form).

(log₂x)(log₃x)(log₄x)(log₅x)  =  [log(x)/log(2)]·[log(x)/log(3)]·[log(x)/log(4)]·[log(x)/log(5)]

                                            =  [log(x)]⁴ / [log(2)·log(3)·log(4)·log(5)] 

(log₂x)(log₃x)(log₄x)  =  [log(x)/log(2)]· [log(x)/log(3)]· [log(x)/log(4)]

                                  =  [log(x)]³ / [log(2)·log(3)·log(4)]

] 

(log₂x)(log₃x)(log₅x)  =  [log(x)/log(2)]· [log(x)/log(3)]· [log(x)/log(5)]

                                  =  [log(x)]³ / [log(2)·log(3)·log(5)]

(log₂x)(log₄x)(log₅x)  =  [log(x)/log(2)]· [log(x)/log(4)]· [log(x)/log(4)]

                                  =  [log(x)]³ / [log(2)·log(4)·log(5)]

(log₃x)(log₄x)(log₅x)  =  [log(x)/log(3)]· [log(x)/log(4)]· [log(x)/log(4)]

                                  =  [log(x)]³ / [log(3)·log(4)·log(5)]

To get rid of denominators, multiply both sides by [log(2)·log(3)·log(4)·log(5):

--->   [log(x)]⁴  =  [log(x)]³ · [log(5)]  +  [log(x)]³ ·[log​(4)]  +  [log(x)]³ ·[log​(3)]  +  [log(x)]³ ·[log​(2)]

                       =  [log(x)]³ · [ log(5) + log(4) + log(3) + log(2) ]

Divide both sides by [log(x)]³ 

--->   log(x)  =  log(5) + log(4) + log(3) + log(2)

                    =  log( 5 · 4 · 3 · 2 )

                    =  log( 120 )

--->  x = 120

I'm sorry but 120 is wrong.