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# SNAP Logarithms Questions PDF

Logarithms is an important topic in the Quant section of the SNAP Exam. You can also download this Free Logarithms Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Logarithms questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

Question 1:Â Find the value of following expression $\log\sin 40^\circ \log \sin 41^\circ — \log \sin 99^\circÂ \log \sin 100^\circ$

a)Â $\frac{\sqrt{3ac}+1}{2}$

b)Â $0$

c)Â $1$

d)Â $2$

Solution:

= $\log\sin 40^\circ \log \sin 41^\circ…..Â \log \sin 90^\circ…… \log \sin 99^\circ \log \sin 100^\circ$

= $\log \sin 90^\circ$

= log 1

= 0

Question 2:Â The value of $\log_2 x$ which satisfy $6 – 9\log_{8}\left(\frac{4}{x}\right)^{\frac{1}{3}} – 8(\log_{256}x)^{\frac{2}{3}} – (\log_2 x^8)^{\frac{1}{3}} = 0$ is

a)Â 2

b)Â $\sqrt{2}$

c)Â 4

d)Â 8

Solution:

$6-9\log_8\left(\frac{4}{x}\right)^{\frac{1}{3}}-8\left(\log_{256}x\right)^{\frac{2}{3}}-\left(\log_{2\ }x^8\right)^{\frac{1}{3}}=0$

$6-\log_24+\log_2\ x-2\left(\log_2x\right)^{\frac{2}{3}}-2\left(\log_2x\right)^{\frac{1}{3}}=0$

LetÂ $\left(\log_2x\right)^{\frac{1}{3}}$ be t.

$4+t^3-2t^2-2t=0$

or,Â $\left(t-2\right)\left(t^2-2\right)=0$

so,Â $t=2$ or

$\log_2\ x=8$

Question 3:Â 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)Â $2<a<3$

b)Â $3<a<4$

c)Â $4<a<5$

d)Â $a>5$

Solution:

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 .

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

Solution:

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

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

Solution:

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

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

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

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

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

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

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

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

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

Question 6:Â If $\log \left(\frac{a}{b}\right) + \log \left(\frac{b}{a}\right) = \log(a + b)$, then which of the following statements is CORRECT?

a)Â a – b = 1

b)Â a + b = 1

c)Â a = b

d)Â $a^2 – b^2 = 1$

Solution:

log(a/b) + log(b/a) = log(a+b)

log(a+b) = log(a/b)(b/a)

log(a+b)=log 1

a+b=1

Question 7:Â 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)Â $\frac{1}{m} + \frac{1}{n} = 1$

b)Â m = n

c)Â $m^2 + n^2 = 1$

d)Â $\frac{1}{m} + \frac{1}{n} = 2$

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

Solution:

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

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

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.

Question 8:Â The value of $\log_{a}({\frac{a}{b}})+\log_{b}({\frac{b}{a}})$, for $1<a\leq b$ cannot be equal to

a)Â 0

b)Â -1

c)Â 1

d)Â -0.5

Solution:

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

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

Now applying the property of AM>=GM, we get thatÂ Â $\frac{\left(\log_ab+\frac{1}{\log_ba}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ba}\right)\ge2$ 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.

Question 9:Â $\frac{2\times4\times8\times16}{(\log_{2}{4})^{2}(\log_{4}{8})^{3}(\log_{8}{16})^{4}}$ equals

Solution:

$\frac{\left(2\cdot4\cdot8\cdot16\right)}{\left(\log_22^2\right)^2\cdot\left(\log_{2^2}2^3\right)^3\cdot\left(\log_{2^3}2^4\right)^4}\cdot$

=Â $\frac{2^{10}}{4\cdot\left(\frac{3}{2}\right)^3\cdot\left(\frac{4}{3}\right)^4}=24$

Question 10:Â If $\log_{a}{30}=A,\log_{a}({\frac{5}{3}})=-B$ and $\log_2{a}=\frac{1}{3}$, then $\log_3{a}$ equals

a)Â $\frac{2}{A+B-3}$

b)Â $\frac{2}{A+B}-3$

c)Â $\frac{A+B}{2}-3$

d)Â $\frac{A+B-3}{2}$

Solution:

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

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

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Question 11:Â If $\log_{4}{5}=(\log_{4}{y})(\log_{6}{\sqrt{5}})$, then y equals

Solution:

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

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

Question 12:Â If Y is a negative number such that $2^{Y^2({\log_{3}{5})}}=5^{\log_{2}{3}}$, then Y equals to:

a)Â $\log_{2}(\frac{1}{5})$

b)Â $\log_{2}(\frac{1}{3})$

c)Â $-\log_{2}(\frac{1}{5})$

d)Â $-\log_{2}(\frac{1}{3})$

Solution:

$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$

Given,Â $5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$

=>Â $Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$

=>$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$

since Y is a negative number, Y=$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$

Question 13:Â Let x and y be positive real numbers such that
$\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3,$ and $\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$. Then $xy$ equals

a)Â 150

b)Â 25

c)Â 100

d)Â 250

Solution:

We have,Â $\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3$

=>Â $x^2-y^2=125$……(1)

$\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$

=>$\ \frac{\ y}{x}$ =Â $\ \frac{\ 2}{3}$

=> 2x=3yÂ  Â => x=$\ \frac{\ 3y}{2}$

On substituting the value of x in 1, we get

$\ \frac{\ 5x^2}{4}$=125

=>y=10, x=15

Hence xy=150

Question 14:Â Sham is trying to solve the expression:
$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circÂ + …….. +Â \log \tan 89^\circ$.

a)Â 1

b)Â $\frac{1}{\sqrt{2}}$

c)Â 0

d)Â -1

Solution:

$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$.

=$\log \tan 1^\circ + \log \tan 89^\circ + \log \tan 2^\circ + \log \tan 88^\circ …….. + \log \tan 45^\circ$.

=$\log\ \left(\tan\ 1^0\cdot\tan\ 89^0\right)\times\log\ \left(\tan\ 2^0\cdot\tan\ 88^0\right)\ ………………………\log\ \left(\tan\ 45^0\right)$

tan $45^0$ = 1

$\log\ \left(\tan\ 45^0\right)\ =\ 0$

$\therefore$Â $\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$ = 0

Question 15:Â If $\log_{10}{11} = a$ then $\log_{10}{\left(\frac{1}{110}\right)}$ is equal to

a)Â $-a$

b)Â $(1 + a)^{-1}$

c)Â $\frac{1}{10 a}$

d)Â $-(a + 1)$

Solution:

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

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

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

= 0$-\log_{10}110$

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

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

= -(a+1)

Question 16:Â Find the value of $\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

a)Â $n^{2} + 1$

b)Â $n^{2} – 1$

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

d)Â $\frac{(n^{2} + n)}{2}$

Solution:

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

SinceÂ $\log_aa\$ = 1

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

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

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

Question 17:Â what is the value of $\frac{\log_{27}{9} \times \log_{16}{64}}{\log_{4}{\sqrt2}}$?

a)Â $\frac{1}{6}$

b)Â $\frac{1}{4}$

c)Â 8

d)Â 4

Solution:

$\frac{\log_{27}{9} \times \log_{16}{64}}{\log_{4}{\sqrt2}}$?

=$\frac{\ \log_{3^3}3^2\times\ \log_{2^4}2^6}{\log_{\left(\sqrt{\ 2}\right)^4}\sqrt{\ 2}}$

=$\frac{\ \ \frac{\ 2}{3}\times\ \frac{\ 6}{4}}{\ \frac{\ 1}{4}}$

=4

Question 18:Â What is the value of x in the following expression?
$x + \log_{10} (1 + 2^x) = x \log_{10} 5 + \log_{10} 6$

a)Â 1

b)Â 0

c)Â -1

d)Â 3

Solution:

The given equation can be written as

$\log\left(10\right)^{x\ }\ +\ \log\left(1+2^x\right)=\log\left(5\right)^x+\log6$

$\log\left(10\right)^{x\ }\left(1+2^x\right)=\log\left(5\right)^x\cdot6$Â  Â  (Â  since logA + logB=logAB)

$\log\ \frac{\left(2^x\cdot5^x\right)\left(1+2^x\right)}{5^x\cdot6}=0$Â  Â Â ( since logA – logB=logA/B)

$\frac{\left(2^x\ +2^{2x\ }\right)}{6}=10^0$Â  ($Since\ \log_aN\ =x\ \ =>N=a^x$)

$2^{^x}+2^{2x}=6$

The aboveÂ Â equation is satisfied only when x=1

Question 19:Â Find the value of $\log_{3^2}{5^4} \times \log_{5^2}{3^4}$

a)Â 5

b)Â 3

c)Â 4

d)Â 2

Solution:

$\log_{b^n}\left(a^m\right)\ =\frac{m}{n}\log_ba\ =\frac{m}{n}\cdot\frac{\log\left(a\right)}{\log\left(b\right)}$

So given equation becomesÂ $\frac{4}{2}\cdot\frac{4}{2}\cdot\frac{\log\left(3\right)}{\log\left(2\right)}\cdot\frac{\log\left(2\right)}{\log\left(3\right)}$Â  = 4

Question 20:Â $\log_{5}{25} + \log_{2} (\log_{3}{81})$ is

a)Â 1

b)Â 2

c)Â 3

d)Â 4

$\log\left(a^m\right)\ =\ m\log\left(a\right)\ and\ \ \log_aa$ = 1
$\log_55^2\ +\ \log_2\left(\log_33^4\right)$
2 +Â $\ \log_24$
2+Â $\ \log_22^2$