Basic math: problem 1

1

\(\frac{(2^3 \times 2^5)^{10}}{512} = \frac{(2^{3+5})^{10}}{2^9} = \frac{2^{80}}{2^9} = 2^{71} \)

2

\((\log_2 13)(\log_{13} 2048) = (\log_2 13)(\log_{13} 2^{11}) = 11 (\log_2 13)(\log_{13} 2)\)

Using the change-of-base formula, we have \((\log_x a)(\log_a b) = \log_x b\), so \((\log_2 13)(\log_{13} 2) = \log_2 2 = 1\)

So \((\log_2 13)(\log_{13} 2048) = 11\)

3

\(\frac{\log_3(81^k)}{7k} = \frac{k \log_3 (3^4)}{7 k} = \frac{4}{7} \)

4

\( (1 + i)(2 - i)(3 - i) = (2 - i + 2i - i^2)(3 - i) = (3 + i)(3 - i) = (9 - i^2) = 9 + 1 = 10 \)