SuicideFuel Problem

[trying to ascend]

[UWSL]Let (a,b,c,d,e) be a geometric progression of ratio a, with a>0 and a different from 1. If the sum of its terms is equal to 13a+12 and x is a positive real number other than 1 such that[/UWSL]

1/loga(x) + 1/logb(x) + 1/logc(x) + 1/logd(x) + 1/loge(x) = 5/2


Therefore x equals?

 

[hateful insul]

if i can rearrange the alphabet i'd put the D in U

 

[TheProphetMuscle]

if i can rearrange the alphabet i'd put the D in U

 

[hateful insul]

chillax , my gf typed it , i didn't type that

 

[NoBitches]

[UWSL]Let (a,b,c,d,e) be a geometric progression of ratio a, with a>0 and a different from 1. If the sum of its terms is equal to 13a+12 and x is a positive real number other than 1 such that[/UWSL]

1/loga(x) + 1/logb(x) + 1/logc(x) + 1/logd(x) + 1/loge(x) = 5/2


Therefore x equals?

I think x=27
a+a^2+a^3+a^4+a^5=13a+12
We use the infinite sum formula
a(1-a^5)/(1-a)=13a+12
a(1-a^5)=(13a+12)(1-a)
a-a^6=13a+12-13a^2-12a
-a^6=-13a^2+12
a^6=13a^2-12
x=a^2
x^3-13x+12=0
(x-1)(x+4)(x-3)=0
x=1,-4,3
a=sqrt(1),sqrt(-4),sqrt(3)
Therefore a can only be postive sqrt(3)
As it must be greater then 0 and cant be 1 (and I dont want it to be imaginary)
Take 1/ log base sqrt(3) (x)=k
then 1/log base sqrt(3)^2 (x)=2k (As the denominator is half)
and ect... 3k,4k ....
so we get 15k=5/2
30k=5
k=1/6
logbasesqrt(3)(x)=6
so x=sqrt(3)^6=27 (Finnaly giving our answer I think)

 

[zephyr]

Wait; I've only just noticed these are pre-fixed as Suifuel.