+0  
 
+7
78
5
avatar+1063 

Find the sum of all solutions to

 

 Apr 3, 2020

Best Answer 

 #1
avatar+20876 
+1

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

 Apr 3, 2020
 #1
avatar+20876 
+1
Best Answer

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

geno3141 Apr 3, 2020
 #2
avatar+111321 
0

That's impressive, geno  !!!!

 

cool cool cool

CPhill  Apr 3, 2020
 #3
avatar+1063 
+7

I'm sorry but 120 is wrong.

 Apr 4, 2020
 #4
avatar+1063 
+7

Your work looked very good, maybe there was a simple calculation error in all of your work.

SpongeBobRules24  Apr 4, 2020
 #5
avatar
0

(log^4(x))/(log(2) log(3) log(4) log(5)) = (log(120) log^3(x))/(log^2(2) log(5) log(9)), solve for x

 

x = 1   and   x = 120

Guest Apr 4, 2020

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