Find the value of x that satisfies the equation
25^-2 = (5^48/x) / ((5^26/x)(25^17/x))
2007-12-15 04:19:27 · 6 answers · asked by ~USA~ 2 in Science & Mathematics ➔ Mathematics
25^-2 = 5^48/x / (5^26/x)(25^17/x)
5^(48/x)/5^(26/x)(5^2)^(17/x) = (5^2)^-2
=> 5^(48/x)/5^(26/x)(5^34/x) = (5)^-4
=> 5^[48/x - (26/x + 34/x)] = 5^(-4)
=>5^(48/x - 60/x) = 5^(-4)
48/x - 60/x = -4
multiply with x
48 - 60 = -4x
-4x = -12
x = 3
2007-12-15 04:33:34 · answer #1 · answered by mohanrao d 7 · 0⤊ 0⤋
First, put everything into the same terms.
since 25=5^2:
5^-4=(5^48/x) / ((5^26/x)(5^34/x))
5^-4=(5^48/x) / (5^60/x)
5^-4 = 5^(48/x-60/x)=5^-12/x
-4=-12/x
x=3.
2007-12-15 12:34:43 · answer #2 · answered by Abby 2 · 0⤊ 0⤋
25^-2 = (5^48/x) / ((5^26/x)(25^17/x))
5^-4 = 5^-12/x
-4 = -12/x
x = -12/-4 = 3
2007-12-15 12:28:48 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
25^8
2007-12-15 12:27:20 · answer #4 · answered by Bradley 1 · 0⤊ 0⤋
-4 = -12/x
x = -12/-4 = 3
2007-12-15 12:39:21 · answer #5 · answered by J 6 · 0⤊ 0⤋
390625
calculators such as the TI-86 rule because they have solver
2007-12-15 12:36:38 · answer #6 · answered by Morgan Mariah 2 · 0⤊ 0⤋