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

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(c) ${a^2} + 4{b^2} = 12ab$$ \Rightarrow $${a^2} + 4{b^2} + 4ab = 16ab$

$ \Rightarrow $${(a + 2b)^2} = 16ab$

$ \Rightarrow $$2\log (a + 2b) = \l

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) ${a^2} + 4{b^2} = 12ab$$ \Rightarrow $${a^2} + 4{b^2} + 4ab = 16ab$

$ \Rightarrow $${(a + 2b)^2} = 16ab$

$ \Rightarrow $$2\log (a + 2b) = \log 16 + \log a + \log b$

$	herefore $ $\log (a + 2b) = {1 \over 2}[\log a + \log b + 4\log 2]$

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

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