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How do you solve #log_x 25=-0.5#?

1 Answer

Dec 1, 2015

# x = 1/625#

Explanation:

Using the property #x^(log_x(a)) = a# we get

#log_x(25) = -0.5#

#=> x^(log_x(25)) = x^(-0.5)#

#=> 25 = 1/sqrt(x)#

#=> sqrt(x) = 1/25#

#=> x = (1/25)^2 = 1/625#

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