`log_2 (1/8)`
To evaluate this, consider the base and argument of the logarithm. The base of the logarithm is 2. And the 8 can be expressed in terms of factor 2. So factoring 8, the expression becomes:
`= log_2 (1/2^3)`
Then, apply the negative exponent rule `a^(-m)=1/a^m` .
`= log_2 (2^(-3))`
To simplify this further, apply the logarithmic rule `log_b (a^m) = m * log_b (a)` .
`= -3 * log_2(2)`
Take note that when the base and argument of a logarithm are the same, the resulting value is 1 `(log_b (b) = 1)` .
`= - 3 * 1`
`= -3`
Therefore, `log_2 (1/8) = -3` .
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