25^3x-1 = 5^5-x
2006-07-31 17:29:32 · 10 answers · asked by zz06 3 in Science & Mathematics ➔ Mathematics
25 = 5^2
Thus,
25^(3x-1) = (5^2)^(3x-1) = 5^(2*(3x-1)) = 5^(6x - 2)
Therefore,
5^(6x-2) = 5^(5-x)
Since the bases are equal, then the exponents are equal.
6x - 2 = 5 - x
Add x to each side.
7x - 2 = 5
Add 2 to each side.
7x = 7
Divide each side by 7.
x = 1
Check:
25^(3*1 - 1) = 5^(5-1)
25^2 = 5^4
625 = 625
2006-07-31 17:33:07 · answer #1 · answered by MsMath 7 · 1⤊ 1⤋
1
2006-08-01 03:02:01 · answer #2 · answered by budweiser 2 · 0⤊ 0⤋
x=1
2006-08-01 00:56:50 · answer #3 · answered by Anonymous · 0⤊ 0⤋
25 = 5^2 , so 25^3x-1 = 5^2^(3x-1)
Therefore, 5^2^(3x-1) = 5^5-x
Multiply the 3x-1 by 2,
5^6x-2 = 5^5-x
Now you can cancel out the 5's, so
6x-2 = 5-x
Move the -x to the left and the -2 to the right side of the equation, (don't forget to change the signs)
6x+x = 5+2
7x = 7
x = 7/7
x = 1
2006-08-01 00:54:29 · answer #4 · answered by Anonymous · 0⤊ 0⤋
25^3x-1 = 5^5-x
5^2(3x-1) = 5^5-x --->(25=5^2, so 5^2(3x-1))
5^6x-2 = 5^5-x
(since you are solving for x just get "6x-2" and "5-x")
6x-2 = 5-x
6x+x = 5+2 (move "x" to the left side and "2" to the right side)
7x = 7
x = 7/7
x = 1
2006-08-01 07:50:10 · answer #5 · answered by Anonymous · 0⤊ 0⤋
25= 5^2
Therfore,
5^(6x - 2) = 5 ^(5-x)
6x - 2 =5 - x
7x = 7
x = 1
2006-08-01 04:49:15 · answer #6 · answered by Niwton 1 · 0⤊ 0⤋
if by this you mean
25^(3x - 1) = 5^(5 - x)
(5^2)^(3x - 1) = 5^(5 - x)
5^(2(3x - 1)) = 5^(5 - x)
5^(6x - 2) = 5^(5 - x)
6x - 2 = 5 - x
7x = 7
x = 1
ANS : x = 1
2006-08-01 11:45:46 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
25^3x-1 = 5^5-x
Or, 5^2(3x-1)=5^(5-x)
Or, 2(3x-1)=5-x
Or, 6x-2=5-x
Or, 7x=7
Or, x=1
2006-08-01 04:00:15 · answer #8 · answered by sharanan 2 · 0⤊ 0⤋
unknowns to the left knowns to the right
gives something times x = somenumber
thus x is somenumber divided by something.
2006-08-01 00:59:23 · answer #9 · answered by gjmb1960 7 · 0⤊ 0⤋
x = 0.0666866
2006-08-01 00:35:48 · answer #10 · answered by Chug-a-Lug 7 · 0⤊ 0⤋