If $${\log _3}x + {\log _{9}}{x^2} + {\log _{27}}{x^3}$$ $$ = 9,$$ then x equals to -
A. 3
B. 9
C. 27
D. None of these
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr} $$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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