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If log a b÷2=1÷2(log a log b) then the value of a^2 b^2?
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If log a b÷2=1÷2(log a log b) then the value of a^2 b^2?
Given Equation:
log a b÷2 = 1÷2(log a - log b)

To find a^2 b^2:
- Step 1: Simplify the given equation
Using the properties of logarithms, we can rewrite the equation as:
log a (b^2) = 1/2 * (log a - log b)^2
- Step 2: Apply the power rule of logarithms
log a (b^2) = 1/2 * (log a / log b)^2
- Step 3: Simplify further
log a (b^2) = 1/2 * (log a)^2 / (log b)^2
- Step 4: Use the definition of logarithms
b^2 = a^(1/2) / b^(1/2)
- Step 5: Square both sides
(b^2)^2 = (a^(1/2) / b^(1/2))^2
b^4 = a/b
- Step 6: Simplify the expression
a^2 * b^2 = b^4 * b^2 = b^6
Therefore, the value of a^2 * b^2 is b^6.
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If log a b÷2=1÷2(log a log b) then the value of a^2 b^2?
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