Use these values and the properties of logarithms to approximate the value.

Question

LogA5=.699 and LogA3=.477 use both or just one to evaluate LogA125=

Answer 1

125=5³

log_A125=3*log_A5=2.097

Answer 2

Since log_A5 = 0.699, log_A125 = log_A(5³) = 3(log_A5) = 3(0.699) = 2.097