How do you evaluate #log_25 5#?

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Answer 1 · 2016-08-21T20:07:06

We have

#log_25 5=log5/(log25)=log5/(log5^2)=log5/(2*log5)=cancellog5/(2*cancellog5)=1/2#

Answer 2 · 2016-08-21T21:16:51

#log_25 5 = 1/2#

You can find the answer by inspection if you understand what question is being asked when you are working in log form.

#log_color(red)(10) color(blue)(100)# asks the question........,

#" What power of "color(red)(10)" will give the number " color(blue)(100)#?"

The answer is obviously 2. Because #10^2 = 100#

#:. log_10 100 = 2 and log_10 1000 = 3#

Can you see that in the same way....

#log_3 27 = 3, and log_5 125 = 3 and log_8 64 = 2 ?#

A square root can also be written as an index.

#sqrtx = x^(1/2) " " and " " sqrt49 = 49^(1/2) =7#

#log_25 5 " asks the question 'what power of 25 will give 5'? "#

As 5 is the square root of 25, the index we need is #1/2#

#log_25 5 = 1/2#

Can you answer these as well?

#log_36 6 ," " log_81 3, " " log_32 2#

#1/2 " " 1/4, " " 1/5#