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Basic of Logarithms
medium
જો ${a^2} + 4{b^2} = 12ab $ તો $\log (a + 2b)= . . .$ .
A
${1 \over 2}[\log a + \log b - \log 2]$
B
$\log {a \over 2} + \log {b \over 2} + \log 2$
C
${1 \over 2}[\log a + \log b + 4\log 2]$
D
${1 \over 2}[\log a - \log b + 4\log 2]$
Solution
(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$
$\therefore $ $\log (a + 2b) = {1 \over 2}[\log a + \log b + 4\log 2]$
Standard 11
Mathematics
