How do you solve #(1/2)^x=32#?

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Answer 1 · 2016-12-29T12:15:18

#x=-5#

#(1/2)^x=32#

there are #2# ways of solving this:

#1)#
convert to logarithmic form:

#a^m=n -> log_a(n)=m#

#(1/2)x=32 -> log_(1/2)(32)=x#

#log_(1/2)(32)=log_0.5(32)#

enter into a calculator:

#log_0.5(32)= -5#

#x=-5#

#2)#

laws of indices:

#a^-m = 1/(a^m)#

#(a^m)^n=a^(mn)#

convert #(1/2)# and #32# to powers of #2#:

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

#32 = 2^5#

#(1/2)^x=32#

#(2^-1)^x=2^5#

#2^(-1 * x) = 2^5#

#-1*x = 5#

#-x = 5#

divide by #-1#:

#x = -5#