A) \[x=2\]
B) \[x=4\]
C) \[x=-4\]
D) \[x=-2\]
Correct Answer: B
Solution :
[b] \[lo{{g}_{7}}lo{{g}_{5}}\left( \sqrt{x+5}+\sqrt{x} \right)=0\] \[\Rightarrow lo{{g}_{5}}\left( \sqrt{x+5}+\sqrt{x} \right)={{7}^{{}^\circ }}=1\Rightarrow \sqrt{x+5}+\sqrt{x}=5'=5\]Squaring both sides, we have \[x+5+x+2\sqrt{x(x+5)}=25\] \[\Rightarrow 2\sqrt{x(x+5)}=20-2x\] \[\Rightarrow \sqrt{x\left( x+5 \right)}=10-x\] Again squaring both sides we have \[{{x}^{2}}+5x=100+{{x}^{2}}-20x\] \[25x=100\Rightarrow x=\frac{100}{25}=4\] Hence, option [b] is correct.
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