Answer:
Step-by-step explanation:
[tex]log\ \frac{1}{x} = - log \ x \\\\log_a \ a = 1\\\\log \ a^x = x \ log \ a[/tex]
(iii)
[tex]m = log_2 \ \frac{1}{32}\\\\m = log_2 \ \ 32^{-1}\\\\m = - 1 \ log_2 32\\\\m = - 1 \ log_2 \ 2^5\\\\m = -1 \times 5 \ log_2 \ 2\\\\m = -5 \ log_2 \ 2\\\\m = -5 \times 1 = -5[/tex]
(iv)
[tex]log_a \ m = 0\\\\m = a^0 = 1[/tex] [tex][ \ log _a \ 1 = 0 \ ][/tex]
(vii)
[tex]log_{32} \ 8 = m\\\\8 = 32^m\\\\8^{ \frac{1}{m}} = 32 \\\\(2^{3 })^ { \frac{1}{m}} = 2^{5}\\\\2^{\frac{3}{m}} = 2^5 \\\\\frac{3}{m} = 5\\\\3 = 5 \times m \\\\m = \frac{3}{5}[/tex]
(viii)
[tex]log_5 \ ( 2m + 5 ) = 3\\\\(2m +5 ) = 5^3\\\\2m + 5 = 125\\\\2m = 125 - 5 \\\\2m = 120 \\\\m = 60[/tex]
(x)
[tex]log_{m^3} \ 64 = \frac{2}{3}\\\\64 = (m^3)^{ \frac{2}{3}}\\\\64 = m^{ 3 \times \frac{2}{3}}\\\\64 = m ^2\\\\\sqrt{64} = m \\\\8 = m[/tex]
