Answer:
Reminder: [tex]log _a x[/tex] answers the question: "what power do I have to raise a to in order to get x?"
Thus said.
[tex]1. \ 64= 2^6 = 4^3 \implies log_464=3\\2. \ 5^1 = 5 \implies log_55 = 1\\3.\ 3^1=3; \ 2^1=2 \implies Log_33 + log_22 = 1+1=2\\4.\ 5^1=5; \ 8^1=8 \implies log_55 - log_88=1-1=0\\5. \ 8=2^3; 32=2^5 \implies log_28+log_232 = 3+5=8[/tex]
Last one, if you already talked about log properties, can be handled as:
[tex]log_28+log_232= log_2(8\times32) = log_2256 =8\ (256=2^8)[/tex]
