+0

+7
78
5
+1063

Find the sum of all solutions to

Apr 3, 2020

#1
+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
+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

geno3141 Apr 3, 2020
#2
+111321
0

That's impressive, geno  !!!!

CPhill  Apr 3, 2020
#3
+1063
+7

I'm sorry but 120 is wrong.

Apr 4, 2020
#4
+1063
+7

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

SpongeBobRules24  Apr 4, 2020
#5
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