Question

Me ayudan por favor con esto

Answer 1

A)
log 200 = log (2³ × 5²)
= log 2³ ⁺ log 5²
= 3 log 2 + 2 log 5
= 3 (0,3010) + 2 (0,6990)
= 2,3010

b)
log 2,5 = log (25/10) = log (5/2)
= log 5 - log 2
= 0,6990 - 0,3010
= 0,3980

c)
log √(0,125) = log (0,125)^(1/2)
= (1/2) log (0,125)
= 0,5 log (125/1000)
= 0,5 log (1/8)
= 0,5 ( log 1 - log 8)
= 0,5 ( 0 - log 2³)
= -(0,5)(2) log 2
= - log 2
= -0,6990