$$\frac{1}{2}\left( {\log x + \log y} \right)$$ will equal to $$\log \left( {\frac{{x + y}}{2}} \right)$$ if -
A. y = 0
B. x = $$\sqrt {\text{y}} $$
C. x = y
D. x = $$\frac{{\text{y}}}{2}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \frac{1}{2}\log \left( {xy} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \log {\left( {xy} \right)^{\frac{1}{2}}} = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow {\left( {xy} \right)^{\frac{1}{2}}} = \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow xy = {\left( {\frac{{x + y}}{2}} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 0 \cr & \Rightarrow {\left( {x - y} \right)^2} = 0 \cr & \Rightarrow x - y = 0 \cr & \Rightarrow x = y\, \cr} $$Related Questions on Logarithm
Which of the following statements is not correct?
A. log10 10 = 1
B. log (2 + 3) = log (2 x 3)
C. log10 1 = 0
D. log (1 + 2 + 3) = log 1 + log 2 + log 3
$${{\log \sqrt 8 } \over {\log 8}}$$ is equal to:
A. $$\frac{1}{6}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{1}{8}$$
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