Question

show that log(5/8)+log(128/125)+log(5/2)=0

Answer 1

2log(5/8)+log(128/125)+log(5/2)=0
2log5-2log8+log128-log125+log5-log2=0
2log5-2log2^3+log2^7-log5^3+log5-log2=0
3log5-4log2-7log2-3log5-3log2=0
3log(5/5)+7log(2/2)=0
3log1+7log1=
log1=0
3*0+7*0=0
0=0
lhs=rhs hence proved

Answer 2

Answer:

Step-by-step explanation:

Log (5/8)2+log(128/125)+log(5/2)

Log(25/64)+ log128/125+log5/2

By logorithmic identities,

Log(25/64)(128/125)(5/2)

(After cancellation)

Log 1

We know,

Log1=o

Therefore,0=0

LHS=RHS

hence proved