Question

Column I		Column II
A	If $2 \log \left(\frac{8}{45}\right)+3 \log \left(\frac{25}{8}\right)-4 \log \left(\frac{5}{6}\right)=\log K$, then the value of $K$ is equal to	P	$\frac{1}{4}$
B	The rational number represented by $\log _{10}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})$, is equal to	Q	$\frac{1}{2}$
C	The digit in the unit's place of the number $17^{1999}+11^{1999}-7^{1999}$, is equal to	R	1
		T	2

• A. (A) T; (B) Q; (C) P
• B. (A) T; (B) Q; (C) Q
• C. (A) T; (B) Q; (C) R
• D. (A) T; (B) Q; (C) S

Answer 1

Answer: (C)

(A)$\text { L.H.S. } =2 \log \left(\frac{8}{45}\right)+3 \log \left(\frac{25}{8}\right)-4 \log \left(\frac{5}{6}\right) $
$ =\log \left(\frac{8^2}{45^2}\right)+\log \left(\frac{25^3}{8^3}\right)-\log \left(\frac{5^4}{6^4}\right)=\log \left(\frac{8^2}{45^2} \times \frac{25^3}{8^3}\right)-\log \left(\frac{5^4}{6^4}\right)$
$ =\log \left(\frac{8^2 \times 25^3 \times 6^4}{45^2 \times 8^3 \times 5^4}\right)=\log 2=\log k \text { (Given) } $
$\therefore k =2 $
(B) $\log _{10}\left(\frac{\sqrt{6-2 \sqrt{5}}+\sqrt{6+2 \sqrt{5}}}{\sqrt{2}}\right)=\log _{10}\left(\frac{(\sqrt{5}-1)+(\sqrt{5}+1)}{\sqrt{2}}\right)=\log _{10}\left(\frac{2 \sqrt{5}}{\sqrt{2}}\right)=\log _{10} \sqrt{10}=\frac{1}{2}$
Alternatively: Let $x =\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}} \Rightarrow x ^2=6+2 \sqrt{4}=10 \Rightarrow x =\sqrt{10}$
$\therefore \log _{10} \sqrt{10}=\frac{1}{2}$ Ans. $]$
(C)$17^{1999}+11^{1999}-7^{1999}$
$7^x \rightarrow \text { last digit is either } 7,9,3,1 $
$17^{1999} \rightarrow \text { last digit is either } 7,9,3,1 $
$7^{1999} \rightarrow \text { last digit is either } 7,9,3,1$
$11^{1999} \rightarrow \text { last digit is either } 1 $
$\therefore \text { Last digit }\left(17^{1999}\right)+1 \text { - last digit }\left(7^{1999}\right)=1 $