How do you solve #(1/2) ^(3x)=8 ^2#?

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Answer 1 · 2016-05-06T01:03:36

I found #x=-2#

We can try by taking #log_2# on both sides and applying some properties:

#log_2(1/2)^(3x)=log_2(8^2)#

#color(red)(3x)[log_2(1)color(blue)(-)log_2(2)]=log_2(64)#

and using the definition of log:

#3x[0-1]=6#

#-3x=6#

#x=-6/3=-2#

Answer 2 · 2016-05-06T17:53:12

Treat as an exponential equation.

#x = -2#

This question is probably easier than it would appear at first. The clue is that 8 is a power of 2.

#(1/2)^(3x) = 8^2#

Make the bases the same.

#(2^(-1))^(3x) = (2^3)^2#

#2^(-3x) = 2^6#

If the bases are equal then the indices are equal.

#-3x = 6# #rArr x = -2#