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ISI2017DCG9
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The solution of $\log_5(\sqrt{x+5}+\sqrt{x})=1$ is
$2$
$4$
$5$
none of these
isi2017dcg
numericalability
logarithms
asked
Sep 18
in
Numerical Ability
by
gatecse
Boss
(
16.8k
points)
recategorized
Nov 15
by
Lakshman Patel RJIT

17
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answer
comment
0
is it d?
0
answer should be $B$. It can be easily check by putting $x=4$
$log_{5}(\sqrt{x+5} + \sqrt{x}) = 1$
$\Rightarrow \sqrt{x+5} + \sqrt{x} = 5 $
$\Rightarrow \sqrt{x+5}  5 = \sqrt{x} $
On squaring both sides,
$\Rightarrow x+5+2510\sqrt{x+5} = x$
$\Rightarrow 30 = 10\sqrt{x+5}$
$\Rightarrow x = 4$
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Answer $B$
Given: $$\log_5(\sqrt{x+5} + \sqrt{x}) = 1$$
Now, Substitute $x = 4$, we get:
$$\log_5(\sqrt{4+5} + \sqrt{4}) = \log_5(\sqrt9+\sqrt4) = \log_55 = 1, \;as\; \log_aa= 1$$
$\therefore \; B$ is the right answer.
answered
Sep 28
by
`JEET
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