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2log(a -b) = log 5 log a log b?
Most Upvoted Answer
2log(a -b) = log 5 log a log b?
Explanation:
To solve this equation, we need to use logarithmic rules. Let's break it down step by step.

Step 1: Simplify the left side of the equation using the power rule of logarithms.
2log(a -b) = log((a-b)^2)

Step 2: Expand the right side of the equation using the product rule of logarithms.
log 5 + log a - log b = log 5a/b

Step 3: Substitute the simplified left side and expanded right side into the original equation.
log((a-b)^2) = log 5a/b

Step 4: Apply the one-to-one property of logarithms to eliminate the logarithm on both sides of the equation.
(a-b)^2 = 5a/b

Step 5: Expand the left side of the equation.
a^2 - 2ab + b^2 = 5a/b

Step 6: Rearrange the equation to isolate one variable.
a^2 - 2ab - (5a/b) + b^2 = 0

Step 7: Use the quadratic formula to solve for the variable.
a = [2b ± sqrt(4b^2 + 20ab - 4b^3)]/2

Step 8: Simplify the solution.
a = b ± sqrt(b^2 + 5ab - b^3)

Therefore, the solution to the equation 2log(a - b) = log 5 + log a - log b is a = b ± sqrt(b^2 + 5ab - b^3).
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