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A³ b³=ab(8-3a-3b),prove that log(a b)/2=1/3(loga logb)?
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A³ b³=ab(8-3a-3b),prove that log(a b)/2=1/3(loga logb)?
Proof:

Given equation: A³ b³=ab(8-3a-3b)

Step 1: Convert the given equation into logarithmic form.
Taking log on both sides, we get:
3logA + 3logB = log(ab) + log(8-3A-3B)

Step 2: Simplify the equation
3logA + 3logB = log(ab) + log(8-3A-3B)
3(logA + logB) = log(ab) + log(8-3A-3B)
3logA logB = log(ab(8-3A-3B))

Step 3: Use the given equation A³ b³=ab(8-3a-3b)
A³ b³/ab = 8-3A-3B
AB = 8-3A-3B
3A + 3B + AB = 8

Step 4: Substitute AB in the equation found in step 2
3logA logB = log(ab(8-3A-3B))
3logA logB = log(ab(3A+3B+AB))
3logA logB = log(ab(3A+3B+8-3A-3B))
3logA logB = log(ab(8))
log(Ab²) = log(8)
Ab² = 8
Ab = √8

Step 5: Simplify Ab² = 8
log(Ab²) = log(8)
logA + 2logB = log8
2logB = log8 - logA
2logB = log(8/A)
logB = 1/2(log(8/A))
log(Ab)/2 = 1/3(logA logB)

Hence, proved.
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A³ b³=ab(8-3a-3b),prove that log(a b)/2=1/3(loga logb)?
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