How do you solve #8^x = .5#?

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Answer 1 · 2016-03-12T12:54:00

We can use the logarithm to find #x=-1/3#

We start by observing that everything in the equation is a power of 2 and rewrite it as such

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

We can then collapse the power by multiplying the exponents

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

Now we can take the #log_2# of both sides giving

#3x=-1#

finally

#x=-1/3#