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Download Logarithms Questions for XAT PDF – XAT Logarithms questions pdf by Cracku. Top 10 very Important Logarithms Questions for XAT based on asked questions in previous exam papers.

Question 1: For a real number a, if $\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$ then a must lie in the range

a) $2<a<3$

b) $3<a<4$

c) $4<a<5$

d) $a>5$

Question 2: If $\log_{2}[3+\log_{3} \left\{4+\log_{4}(x-1) \right\}]-2=0$ then 4x equals

Question 3: If $5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$, then 100x equals

Question 4: If $\log_4m + \log_4n = \log_2(m + n)$ where m and n are positive real numbers, then which of the following must be true?

a) $\frac{1}{m} + \frac{1}{n} = 1$

b) m = n

c) $m^2 + n^2 = 1$

d) $\frac{1}{m} + \frac{1}{n} = 2$

e) No values of m and n can satisfy the given equation

Question 5: The value of $\log_{a}({\frac{a}{b}})+\log_{b}({\frac{b}{a}})$, for $1<a\leq b$ cannot be equal to

a) 0

b) -1

c) 1

d) -0.5

Question 6: If $\log_{a}{30}=A,\log_{a}({\frac{5}{3}})=-B$ and $\log_2{a}=\frac{1}{3}$, then $\log_3{a}$ equals

a) $\frac{2}{A+B-3}$

b) $\frac{2}{A+B}-3$

c) $\frac{A+B}{2}-3$

d) $\frac{A+B-3}{2}$

Question 7: If $\log_{4}{5}=(\log_{4}{y})(\log_{6}{\sqrt{5}})$, then y equals

Question 8: If $\log_{10}{11} = a$ then $\log_{10}{\left(\frac{1}{110}\right)}$ is equal to

a) $-a$

b) $(1 + a)^{-1}$

c) $\frac{1}{10 a}$

d) $-(a + 1)$

Question 9: Find the value of $\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

a) $n^{2} + 1$

b) $n^{2} – 1$

c) $\frac{(n^{2} + n)}{2}.\frac{n(n + 1)}{3}$

d) $\frac{(n^{2} + n)}{2}$

Question 10: what is the value of $\frac{\log_{27}{9} \times \log_{16}{64}}{\log_{4}{\sqrt2}}$?

a) $\frac{1}{6}$

b) $\frac{1}{4}$

c) 8

d) 4

We have :$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$
We get $\frac{\left(\frac{\log a}{\log\ 15}+\frac{\log a}{\log32}\right)}{\frac{\log a}{\log\ 15}\times\ \frac{\log a}{\log32}\ \ }=4$
we get $\log a\left(\log32\ +\log\ 15\right)=4\left(\log\ a\right)^2$
we get $\left(\log32\ +\log\ 15\right)=4\log a$
=$\log480=\log a^4$
=$a^4\ =480$
so we can say a is between 4 and 5 .

We have :
$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$
we get $3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$
we get $\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$
we get $4+\log_4\left(x-1\right)\ =\ 3$
$\log_4\left(x-1\right)\ =\ -1$
x-1 = 4^-1
x = $\frac{1}{4}+1=\frac{5}{4}$
4x = 5

$5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$

We can re-write the equation as: $5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$

$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$

$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$

$5=-5\log_{10}\sqrt{1-x}$

$\sqrt{1-x}=\frac{1}{10}$

Squaring both sides: $\left(\sqrt{1-x}\right)^2=\frac{1}{100}$

$\therefore\$ $x=1-\frac{1}{100}=\frac{99}{100}$

Hence, $100\ x\ =100\times\ \frac{99}{100}=99$

$\log_4mn=\log_2(m+n)$

$\sqrt{\ mn}=(m+n)$

Squarring on both sides

$m^2+n^2+mn\ =\ 0$

Since m, n are positive real numbers, no value of m and n satisfy the above equations.

On expanding the expression we get $1-\log_ab+1-\log_ba$

$or\ 2-\left(\log_ab+\frac{1}{\log_ba}\right)$

Now applying the property of AM>=GM, we get that  $\frac{\left(\log_ab+\frac{1}{\log_ba}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ba}\right)\ge2$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.

$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$………..(1)

$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$………….(2)

and finally $\log_a2=3$

Substituting this in (1) we get $\log_a5+\log_a3=A-3$

Now we have two equations in two variables (1) and (2) . On solving we get

$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$

$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$

$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$

$\log_{10}{\left(\frac{1}{110}\right)}$

$\log_a\left(\ \frac{\ x}{y}\right)\ =\ \log_ax-\log_ay$

$\log_{10}{\left(\frac{1}{110}\right)}$ = $=\ \log_{10}1-\log_{10}110$

= 0$-\log_{10}110$

=$-\log_{10}11\times\ 10$

=$-\left(\log_{10}11+\log_{10}10\right)$

= -(a+1)

$\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

Since $\log_aa\$ = 1

$\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$ = 1+2+….n

=$\ \frac{\ n\left(n+1\right)}{2}$

=$\frac{(n^{2} + n)}{2}$

$\frac{\log_{27}{9} \times \log_{16}{64}}{\log_{4}{\sqrt2}}$?
=$\frac{\ \log_{3^3}3^2\times\ \log_{2^4}2^6}{\log_{\left(\sqrt{\ 2}\right)^4}\sqrt{\ 2}}$
=$\frac{\ \ \frac{\ 2}{3}\times\ \frac{\ 6}{4}}{\ \frac{\ 1}{4}}$