How do you solve $8^{x} = .5$ $8^x = .5$ ?

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Answer 1 · 2016-03-12T12:54:00

We can use the logarithm to find $x = - \frac{1}{3}$ $x=-1/3$

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

${(2^{3})}^{x} = 2^{- 1}$ $(2^3)^x=2^-1$

We can then collapse the power by multiplying the exponents

$2^{3 x} = 2^{- 1}$ $2^(3x)=2^-1$

Now we can take the ${log}_{2}$ $log_2$ of both sides giving

$3 x = - 1$ $3x=-1$

finally

$x = - \frac{1}{3}$ $x=-1/3$

How do you solve 8x=.58^x = .5?