Question

can you plss solve both the parts

Answer 1

1:
$\log_{3}(9) + \log_{3}(3) \\ = \log_{3}( {3}^{2} ) + \log_{3}(3) \\ = 2 \log_{3}(3) + \log_{3}(3) \\ = 2 + 1 \\ = 3 \\ \\ \log_{5}(125) \\ = \log_{5}( {5}^{3} ) \\ = 3 \log_{5}(5) \\ = 3 \\ \text{so we can conclude} \log_{3}(9) + \log_{ 3}(3) = \log_{5}(125)$

2:

The base is not given but I am going to assume base 10(just for the reason it gives pretty simple answer).

$3 \log_{10}(5) - \log_{10}(50) + 2 \log_{10}(2) + 1 \\ = \log_{10}(5 ^{3} ) + \log_{10}(50 ^{ - 1} ) + \log_{10}(2 ^{2} ) + \log_{10}(10) \\ = \log_{10}( {5}^{3} \times \frac{1}{50} \times {2}^{2} \times 10 ) \\ = \log_{10}(100) \\ = \log_{10}( {10}^{2} ) \\ = 2 \log_{10}(10) \\ = 2$