Find the value of \(x\) that satifies the equation \(25^{-2}= \frac{5^{48/x}}{5^{26/x}\cdot 25^{17/x}}\)
Solve for x over the real numbers:
1/625 = 5^(-12/x)
1/625 = 5^(-12/x) is equivalent to 5^(-12/x) = 1/625:
5^(-12/x) = 1/625
Take reciprocals of both sides:
5^(12/x) = 625
Take the logarithm base 5 of both sides:
12/x = 4
Take the reciprocal of both sides:
x/12 = 1/4
Multiply both sides by 12:
x = 3
+128408 Here's an approach where logs aren't required
25^(-2) = (5^2)^(-2) = 5^(-4)
25^(17/x) = (5^2)^(17/x) = 5^(34/ x)
So we have
5^(-4) = 5^(48/x)
_______________
5^(26/x) * 5^(34/x)
5^ (-4) = 5^(48/x)
_________
5^(60/x)
5^(-4) = 5^(-12/ x)
Solve for the exponents
-4 = -12 / x
-4x = - 12
x = -12 / - 4 = 3
