How do you evaluate #log_625 25#?

2 Answers
Sep 1, 2016

#log_625 25 = color(green)(1/2)#

Explanation:

Note that #625 = 25^2color(white)("XX")rarr 25 = 625^(1/2)#

Therefore
#color(white)("XXX")log_625 25 = log_625 626^(1/2)#

#color(white)("XXX")rarr log_625 25 = 1/2# (by definition of log)

Sep 1, 2016

#log_625 25 = 1/2#

Explanation:

Logs are easier to understand if you think about an expression given in log form as asking a question.

In #log_625 25#, the question being asked is;

"What power/index of 625 will give 25?"
OR " How do I make 625 into 25?"

You should recognise #25# as being the square root of 625.

A square root can be written as an index.

#sqrtx = x^(1/2)#

#sqrt625 = 625^(1/2) larr # this answers our question!

#log_625 25 = 1/2#

Log form and index form are interchangeable.

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

#log_625 25 = x hArr 625^x = 25#

#625^x = (25^2)^x= 25^1 larr # make the bases the same

#25^(2x) = 25^1 larr # equate the indices and solve for x

#2x = 1 rarr x = 1/2#