Question

11. log125/27=
P(log5-log 3) then find the value of P3P+1
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Answer 1

log(125/27)

= log(5/3)^3

= 3log(5/3)

= 3 ( log5 - log3)

log(125/27) = p(log5-log3)

3 (log5-log3) = p(log5-log3)

by comparing, p = 3.

put the value of p in the equation