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

Question

↳Redirected from "What are molarity, molality, and normality?"

5279 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-21T05:50:33

#x=4#

#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#