How do you solve for x in #log_2x=log_4(25)#?

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Answer 1 · 2018-05-01T20:48:05

#x=5#

Using the formal definiton of a logarithm, we can take the right hand side to be:

#log_4 25=a=>4^a = 25#

Since #4# is simply #2^2#, we have:

#2^(2a)=25#

Remember; #a# is equal to #log_4 25#, which is in turn equal to #log_2 x#.

#2^(2log_2x)=25#

Taking the binary logarithm of both sides:

#2log_2x=log_2 25#

#=> log_2x = 1/2log_2 25#

Knowing that #alog_b c = log_b c^a#, we reach the result we wished to get:

#log_2x = log_2 (25)^(1/2) = log_2 5 => color(red)(x=5)#