2015
Mathematics
67ca00ce0c643c71d77dee1f
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−3−1/2log5THIS WEEK's
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