Question

I really need your help. I am having a lot of difficulty in solving this question, so please help me..​

Answer 1

Explanation:

`log\ (1)/(x) = - log \ x \\\\log_a \ a = 1\\\\log \ a^x = x \ log \ a`

(iii)

`m = log_2 \ (1)/(32)\\\\m = log_2 \ \ 32^(-1)\\\\m = - 1 \ log_2 32\\\\m = - 1 \ log_2 \ 2^5\\\\m = -1 * 5 \ log_2 \ 2\\\\m = -5 \ log_2 \ 2\\\\m = -5 * 1 = -5`

(iv)

`log_a \ m = 0\\\\m = a^0 = 1`
`[ \ log _a \ 1 = 0 \ ]`

(vii)

`log_(32) \ 8 = m\\\\8 = 32^m\\\\8^{ (1)/(m)} = 32 \\\\(2^(3 ))^ { (1)/(m)} = 2^(5)\\\\2^{(3)/(m)} = 2^5 \\\\(3)/(m) = 5\\\\3 = 5 * m \\\\m = (3)/(5)`

(viii)

`log_5 \ ( 2m + 5 ) = 3\\\\(2m +5 ) = 5^3\\\\2m + 5 = 125\\\\2m = 125 - 5 \\\\2m = 120 \\\\m = 60`

(x)

`log_(m^3) \ 64 = (2)/(3)\\\\64 = (m^3)^{ (2)/(3)}\\\\64 = m^{ 3 * (2)/(3)}\\\\64 = m ^2\\\\√(64) = m \\\\8 = m`