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³log2² = b
2 (³log2) = b
³log2 = b/2
dg menggunakan inverse, maka ²log3 = 2/b
maka :
a) ³log72 = ³log8 + ³log9 = ³log2³ + 2 = 3(³log2) + 2 = 3(b/2) + 2
= 3/2 b
b) ²log36 = ²log4+²log9 =2+ ²log3² = 2 + 2 (²log3) = 2 + 2(2/b)
= 2 + 4/b
c) 6' log 64 = ²log64 / ²log6 = 6 / (²log2 + ²log3) = 6/(1 + 2/b)
= 6/(1 + 2/b) x b/b = 6b/(b+2)
d) 4' log 54 = ²log54/²log4 = (²log2 + ²log27)/2
= (1 + ²log3³)/2 = (1 + 3(²log3))/2 = (1 + 3(2/b))/2
= (1 + 6/b)/2 = (b + 6)/(2b)
a.³log72=³log4+³log3+³log6=b+1+³log6
b.²log36=2.²log6=2.(²log2+1/³log2)=2.(1+1/b/2)=2+(2.1/b/2)
=2+4/b
c.6log64=²log64/²log6=6/(1+1/³log2)=6/(1+2/b)
d.4log54=4log9+4log6=3/2.²log3+1/2.(1+²log3)=3/2.2/b+1/2+1/2.2/b=6/2b+1/2+b=