If a^2 + 4b^2 = 12ab,then log(a + 2b) = | vedclass

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(C) Given the equation $a^2 + 4b^2 = 12ab$. Adding $4ab$ to both sides,we get $a^2 + 4b^2 + 4ab = 12ab + 4ab$. $(a + 2b)^2 = 16ab$. Taking the logarit

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$\frac{1}{2}[\log a + \log b + 4\log 2]$

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(C) Given the equation $a^2 + 4b^2 = 12ab$. Adding $4ab$ to both sides,we get $a^2 + 4b^2 + 4ab = 12ab + 4ab$. $(a + 2b)^2 = 16ab$. Taking the logarithm on both sides: $\log(a + 2b)^2 = \log(16ab)$. $2\log(a + 2b) = \log 16 + \log a + \log b$. $2\log(a + 2b) = \log(2^4) + \log a + \log b$. $2\log(a + 2b) = 4\log 2 + \log a + \log b$. Dividing by $2$,we get $\log(a + 2b) = \frac{1}{2}[\log a + \log b + 4\log 2]$.

If $a^2 + 4b^2 = 12ab$,then $\log(a + 2b) =$

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