if 2^a=3^b=6^c then show that c=ab/a+b
Answers
Answer:
2^a=3^b=6^c
Taking log we get
log2^a=log3^b=log6^c
Let it be k
Therefore,log2^a=log3^b=log6^c=k
We know that loga^x = x*loga
Therefore we can write the above equation has
a*log2=b*log3=c*log6=k
Now,
a*log2=k (1)
b*log3=k (2)
c*log6=k (3)
From (1), (2), (3) we get
a=k/log2 , log2=k/a
b=k/log3 , log3=k/b
c=k/log6 (6)
Here 6 can be written has 2*3
There fore log6=log(2*3)
So, c=k/log(2*3)
We know that log(a*b)=loga + logb
Therefore c=k/log2+log3
From (4),(5)
c= k
[(k/a)+(k/b)]
c= k
k(1/a+1/b) (Here K gets cancelled)
Therefore,
c= 1
1/a+1/b
c= 1
a+b
a*b
c= ab
a+b
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Answer:
Answer:
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