How do you solve #log_2(24) - log_2 3 = log_5(x)#?

1 Answer
Alan P.
Dec 21, 2015

#x=125#

Explanation:

#log_2(24)-log_2(3)= log_2(24/3) = log_2(8) = 3#
#color(white)("XXXXXXXXXXXXXXXXXXXXXXXX")#since 2^3=8#

Therefore
#color(white)("XXX")log_2(24)-log_2(3)=log_5(x)#
#rarr#
#color(white)("XXX")log_5(x) = 3#
#rarr#
#color(white)("XXX")x = 5^3 =125#

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