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)$17^{1999}+11^{1999}-7^{1999}$
$7^x\to \text{ last digit is either }7,9,3,1$
$17^{1999}\to \text{ last digit is either }7,9,3,1$
$7^{1999}\to \text{ last digit is either }7,9,3,1$
$11^{1999}\to \text{ last digit is either }1$
$\therefore \text{ Last digit }\left(17^{1999}\right)+1\text{ - last digit }\left(7^{1999}\right)=1$