How do you evaluate #log_(1/6) (1/36)#?
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1 Answer
Nov 4, 2016
Explanation:
Using the
#color(blue)"law of logarithms"#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(log_b x=nhArrx=b^n)color(white)(2/2)|)))#
#rArr1/36=(1/6)^n=1/6^n#
#rArr1/6^2=1/6^nrArrn=2" is the value"#
