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How do you solve for #log_x (1/64) = -3#?

1 Answer

Nov 21, 2015

#x=4#

Explanation:

#log_x(1/64)=-3#

#x^(log_x(1/64))=x^-3#

#1/64=x^-3#

#(1/64)^(-1/3)=(x^-3)^(-1/3)#

#(64)^(1/3)=x#

#4=x#

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