If a2+b2=7ab, prove that log((a+b)/3)=1/2(log a + log b)

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ScanSolve AI Tutor · Accepted Answer

Here's the solution to the problem. Problem: If a² + b² = 7ab, prove that log ((a+b)/(3)) = 1/2(log a + log b). Step 1: Start with the given equation and add 2ab to both sides to form a perfect square. a² + b² = 7ab; a² + b² + 2ab = 7ab + 2ab; (a+b)² = 9ab Step 2: Take the square root of both sides. We assume a, b > 0 for the logarithms to be defined, which implies a+b > 0. √((a+b)²) = √(9ab); a+b = 3√(ab) Step 3: Divide both sides by 3. (a+b)/(3) = √(ab) Step 4: Take the logarithm of both sides. log ((a+b)/(3)) = log (√(ab)); Step 5: Apply the logarithm property log(xⁿ) = n log x. log ((a+b)/(3)) = log ((ab)^(1/2)); log ((a+b)/(3)) = 1/2 log (ab); Step 6: Apply the logarithm property log(xy) = log x + log y. log ((a+b)/(3)) = 1/2 (log a + log b) This completes the proof. Proven: log ((a+b)/(3)) = 1/2(log a + log b)

If a2+b2=7ab, prove that log((a+b)/3)=1/2(log a + log b)

Step-by-step solution

Start with the given equation and add $2ab$ to both sides to form a perfect square.

Take the square root of both sides.

Divide both sides by 3.

Take the logarithm of both sides.

Apply the logarithm property $\log(x^n) = n \log x$ .

Apply the logarithm property $\log(xy) = \log x + \log y$ .

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More Mathematics Questions

If a2+b2=7ab, prove that log((a+b)/3)=1/2(log a + log b)

Step-by-step solution

Start with the given equation and add $2ab$ to both sides to form a perfect square.

Take the square root of both sides.

Divide both sides by 3.

Take the logarithm of both sides.

Apply the logarithm property log⁡(xn)=nlog⁡x\log(x^n) = n \log xlog(xn)=nlogx.

Apply the logarithm property log⁡(xy)=log⁡x+log⁡y\log(xy) = \log x + \log ylog(xy)=logx+logy.

Still stuck on this one?

More Mathematics Questions

Apply the logarithm property $\log(x^n) = n \log x$ .

Apply the logarithm property $\log(xy) = \log x + \log y$ .