Solve the equation & find out its value : 4*(9/4) ^ x - 8*(3/2) ^ (x+1) + 9 = 0.
2006-12-01 21:49:19 · 5 answers · asked by rakesh r 1 in Science & Mathematics ➔ Mathematics
Hmm! (9/4)^x=(3/2)^2x. Now, let (3/2)^x equal some k. equation transforms to 9k^2-8k*3/2+9=0
-->4k^2-12k+9=0
-->(2k-3)^2=0
-->2k-3=0
-->k=3/2
but k =(3/2)^x
therefore x=1
:)
2006-12-01 21:56:38 · answer #1 · answered by netsavvy_sashi 2 · 0⤊ 0⤋
x=1
2006-12-01 22:07:21 · answer #2 · answered by Selphie 3 · 0⤊ 0⤋
The answer is 1
9/4=(3/2)^2
(3/2)^(x+1)= (3/2)^x. (3/2)
Denote (3/2)^x by t
Then
4t^2-12t+9=0
(2t-3)^2=0
2t-3=0
2t=3
t=3/2
t=(3/2)^x=(3/2)^1
the bases are same
Therefore x=1
2006-12-01 21:58:33 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
4(9/4)^x - 8(3/2)^(x+1) + 9 = 0
4[(3/2)^x]^2-8[(3/2)^x]*(3/2)+9=0
4[(3/2)^x]^2-12[(3/2)^x]+9=0
Let k=(3/2)^x
4k^2-12k+9=0
(2k-3)^2=0
k=3/2
(3/2)^x=3/2
x=1
2006-12-01 22:52:13 · answer #4 · answered by Ranna Renni 2 · 0⤊ 0⤋
put(3/2)=t
4t^2x-8t^(x+1)+9=0
4t^2x-64t^x+9=0
use the quadratic formula to solve for t^x
and then revert back to t
2006-12-01 22:24:31 · answer #5 · answered by raj 7 · 0⤊ 0⤋