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

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

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

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

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

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