How do you solve #log_m 2=4#?

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Answer 1 · 2016-09-11T12:10:54

#m = root4 2 = 2^(1/4)#

I made the assumption that the question as #log-m+2 =4# was meant to be #log_m 2 =4#?

Log form and index from are interchangeable. It is often easier to understand and do the question by changing the format.

#log_a b= c " "hArr " " a^c = b#

#log_m 2 =4 = m^4 = 2#

#root4 m^4 = root4 2#

#m = root4 2 = 2^(1/4)#

While there are always 2 possible roots, the negative root does not work, leaving us with only one possible solution for m.

Thank you, Tazwar Sikder for the insight. :)