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2015
Mathematics
JAMB
Evaluate log5(y
2
x
5
÷125b½)
A.
2 log
5
y + 5log5 y
2
− 3
B.
log
5
y
2
+ 5log
5
x + 3
C.
25logy
5
+ 3
D.
2log
5
y + 5log
5
x − ½ log
5
b −3
Correct Answer:
2log
5
y + 5log
5
x − ½ log
5
b −3
Explanation
log
5
(y
2
x
5
÷125b(1/2)) = log
5
y
2
+ log
5
x
5
−[log
5
125+log
5
(1/2) = 2log
5
y + 5log
5
x − log
5
5
3
−1/2log
5
b = 2log
5
y+5log
5
x−3−1/2log
5
log
5
(y
2
x
5
÷125b(1/2)) = log
5
y
2
+ log
5
x
5
−[log
5
125+log
5
(1/2) = 2log
5
y + 5log
5
x − log
5
5
3
−1/2log
5
b = 2log
5
y+5log
5
x−3−1/2log
5
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