Use logs to solve #2x^(-1/4)=4(1/4)# ?

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Answer 1 · 2018-05-20T14:35:19

#x=16#

No need for logarithms here...

Given: #2x^(-1/4)=4(1/4)#.

#=>2x^(-1/4)=1#

#=>x^(-1/4)=1/2#

#=>1/(x^(1/4))=1/2#

#=>x^(1/4)=2#

#=>x=2^4=16#

Answer 2 · 2018-05-20T18:18:23

#x=16#

You can simplify before you need to use logs.

#2x^(-1/4) = 4(1/4)" "larr# divide both sides by #2#

#x^(-1/4) = 2(1/4) = 1/2#

Invert to get rid of the negative index.

#x^(1/4) = 2" "larr# log both sides

#log x^(1/4) = log 2#

#1/4log x = log2" "larr# multiply both sides by #4#

#logx = 4log2#

#logx = log2^4" "larr# drop the logs

#x = 2^4#

#x=16#