Question

2log5 + 2log6 + log15 + log2 - log9
pls solve this log​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ 2. log(5) + 2. log(6) + log(15) + log(2) - log(9) \\ \\ we \: have \\ m. log(a) = log(a {}^{m} ) \\ \\ = log(5 {}^{2} ) + log(6 {}^{2} ) + log(15) + log(2) - log(9) \\ \\ = log(25) + log(36) + log(15) + log(2) - log(9) \\ \\ we \: know \: that \\ \\ log(a) + log(b) + log(c) + log(d) = log(abcd) \\ \\ = log(36 \times 25 \times 15 \times 2) - log(9) \\ \\ we \: have \: \\ log(a) - log(b) = log( \frac{a}{b} ) \\ \\ = log( \frac{36 \times 25 \times 15 \times 2}{9} ) \\ \\ = log(4 \times 25 \times 15 \times 2) \\ \\ = log(100 \times 30) \\ \\ = log(1000 \times 3) \\ \\ = log(1000) + log(3) \\ \\ = log(10 {}^{3} ) + log(3) \\ \\ = 3.log(10) + log(3) \\ \\ = 3(1) + log(3) \\ \\ = 3 + log(3)$