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How do you solve #-5 = log_2 x#?

1 Answer

Oct 12, 2015

#x=1/32#

Explanation:

#log_2 x=-5#

Using the rule #log_a M=N rArr M=a^N#, you can rewrite #log_2x=-5# as #x=2^-5#

Using the indices rule #a^-m=1/a^m#,

#x=1/2^5#

#x=1/32#

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