Question

log108/605 = 2 log2 + 3log3-log5-2log11 prove it​

Answer 1

Step-by-step explanation:

$log\frac{108}{605}$

Explanation:

2log2+3log3-log5-2log11

= log2²+log3³-[log5+log(11)²]

$\boxed {loga^{n} = nloga}$

= (log4+log27)-[log5+log121]

= log(4×27)-log(5×121)

$\boxed {logM+logN=logMN}$

= log (108)-log(605)

$= log\frac{108}{605} \: proved$

Answer 2

Answer:

up one is the correct answer