Michael was asked to properly evaluate this logarithm. Which answer should he come to? \( \log _{9} 1 / 01 \) -2 \( -1 / 2 \) -9 \( 1 / 2 \)

To evaluate \( \log_{9} \left( \frac{1}{01} \right) \), let’s first rewrite \( \frac{1}{01} \) as \( \frac{1}{1} = 1 \). The logarithm \( \log_{b}(1) \) is always 0 for any base \( b \), including 9. So, Michael should find that the result is 0. However, if he were dealing with a different number for evaluation, it would be helpful to remember that logarithmic identities, such as \( \log_{b}(a \cdot c) = \log_{b}(a) + \log_{b}(c) \), are handy for breaking down complex expressions. Just don't forget that the logarithm of 1, no matter the base, is always a zero! When it comes to common mistakes, a frequent error is confusing the base or miscalculating the antilogarithm. A careful check of each part can save one from that tricky trap. For instance, if the logarithm's argument were wrongly identified, it would drastically change the outcome. Always make sure the logarithmic function is operating on the intended number!