passpadi

2015

Mathematics

JAMB

Evaluate log5(y2x5÷125b½)

A.

2 log5y + 5log5 y2 − 3

B.

log5 y2 + 5log5 x + 3

C.

25logy 5 + 3

D.

2log5y + 5log5x − ½ log5b −3

Correct Answer: 2log5y + 5log5x − ½ log5b −3

Explanation

log5(y2x5÷125b(1/2)) = log5y2 + log5x5−[log5125+log5(1/2) = 2log5y + 5log5x − log553−1/2log5b = 2log5y+5log5x−3−1/2log5

log5(y2x5÷125b(1/2)) = log5y2 + log5x5−[log5125+log5(1/2) = 2log5y + 5log5x − log553−1/2log5b = 2log5y+5log5x−3−1/2log5
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