How do you evaluate #\log _ { 16} \frac { 1} { 32}#?

1 Answer
Jim G.
May 9, 2018

#-5/4#

Explanation:

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)log_b x=nhArrx=b^n#

#"let "log_(16)( 1/32)=n#

#rArr1/32=16^n#

#"now "32=2^5" and "16=2^4#

#rArr2^-5=2^(4n)#

#"equating the exponents"#

#4n=-5rArrn=-5/4#

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