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Find the value of \(x\) that satifies the equation \(25^{-2}= \frac{5^{48/x}}{5^{26/x}\cdot 25^{17/x}}\)

 Jan 8, 2019
 #1
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+1

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

 Jan 8, 2019
 #2
avatar+103138 
+1

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

 

 

cool cool cool

 Jan 8, 2019

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