Which of the following are equivalent to \( 7 \cdot \log \left(u^{-1 / 5}\right) \) ? Assume \( u>0 \). \( \square \log \left(\frac{1}{\sqrt[5]{u^{7}}}\right) \) \( -\frac{7}{5} \log (u) \) \( -\log \left(u^{7 / 5}\right) \) \( \square \log \left(\frac{1}{u^{7 / 5}}\right) \) \( \square-\frac{1}{5} \log \left(u^{7}\right) \)

To simplify \( 7 \cdot \log \left(u^{-1/5}\right) \), we first apply the property of logarithms, which states that \(\log(a^b) = b \cdot \log(a)\). This gives us: \[ 7 \cdot \log \left(u^{-1/5}\right) = 7 \cdot \left(-\frac{1}{5} \cdot \log(u)\right) = -\frac{7}{5} \cdot \log(u). \] So, the expression is equivalent to \( -\frac{7}{5} \log (u) \). Additionally, using another property of logarithms, we can rewrite it using the identity \(\log\left(\frac{1}{x}\right) = -\log(x)\): \[ 7 \cdot \log \left(u^{-1/5}\right) = -7 \cdot \log\left(\sqrt[5]{u}\right) = -\log\left(u^{7/5}\right). \] Putting all these together, we conclude: 1. The second option, \( -\frac{7}{5} \log (u) \), is equivalent. 2. The third option, \( -\log \left(u^{7/5}\right) \), is also equivalent. For the other statements: - \( \log \left(\frac{1}{\sqrt[5]{u^{7}}}\right) = -\frac{1}{5}\log \left(u^{7}\right) \), but when expanding doesn't lead to the required simplification, so it’s not equivalent. - None of the other proposed forms match the established equivalence, leaving \( -\frac{7}{5} \log (u) \) and \( -\log \left(u^{7/5}\right) \) as the valid transformations of the original expression. Hence, we found that: - \( -\frac{7}{5} \log (u) \) is indeed equivalent. - \( -\log \left(u^{7/5}\right) \) is also equivalent.