If $${\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right)$$ $$ = 0,$$ what is the value of x ?
A. 2
B. 3
C. 4
D. 5
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0 \cr & \Rightarrow {\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = {7^0} = 1 \cr & \Rightarrow \sqrt {x + 5} + \sqrt x = {5^1} = 5 \cr & \Rightarrow {\left( {\sqrt {x + 5} + \sqrt x } \right)^2} = 25 \cr & \Rightarrow \left( {x + 5} \right) + x + 2\sqrt {x + 5} \sqrt x = 25 \cr & \Rightarrow 2x + 2\sqrt{ x}\,\, \sqrt {x + 5} = 20 \cr & \Rightarrow \sqrt {x}\,\, \sqrt {x + 5} = 10 - x \cr & \Rightarrow x\left( {x + 5} \right) = {\left( {10 - x} \right)^2} \cr & \Rightarrow {x^2} + 5x = 100 + {x^2} - 20x \cr & \Rightarrow 25x = 100 \cr & \Rightarrow x = 4 \cr} $$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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