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# CAT Logarithm Questions PDF [Most Important]

The logarithm is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Logarithm questions in CAT Previous year papers. If you want to learn the basics, you can watch these videos on Logarithm basics. This article will look into some important Logarithm Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Logarithm Most Important Questions PDF below, which is completely Free.

Question 1: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 2: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 3: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 4: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 5: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

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

Checkout: CAT Free Practice Questions and Videos

Question 6: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 7: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 8: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 9:Â If p$^{3}$ = q$^{4}$ = r$^{5}$ = s$^{6}$, then the value of $log_{s}{(pqr)}$ is equal to

a)Â $\frac{47}{10}$

b)Â $\frac{24}{5}$

c)Â $\frac{16}{5}$

d)Â $1$

Solution:

Given that,Â p$^{3}$ = q$^{4}$ = r$^{5}$ = s$^{6}$

p$^{3}$=s$^{6}$

p =Â s$^{\frac{6}{3}}$ = s$^{2}$Â  Â …(1)

Similarly,Â q =Â s$^{\frac{6}{4}}$ =Â s$^{\frac{3}{2}}$Â  Â …(2)

Similarly, r =Â s$^{\frac{6}{5}}$Â  Â …(3)

$\Rightarrow$ $log_{s}{(pqr)}$

By substituting value of p, q, and r from equation (1), (2) and (3)

$\Rightarrow$ $log_{s}{(s^{2}*s^{\frac{3}{2}}*s^{\frac{6}{5}})}$

$\Rightarrow$ $log_{s}(s^{\frac{47}{10}})$

$\Rightarrow$ $\dfrac{47}{10}$

Hence, option A is the correct answer.

Question 10:Â $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$=?

a)Â $\frac{1}{2}$

b)Â 10

c)Â 0

d)Â âˆ’4

Solution:

We know thatÂ $\dfrac{1}{log_{a}{b}}$ = $\dfrac{log_{x}{a}}{log_{x}{b}}$

Therefore, we can say thatÂ $\dfrac{1}{log_{2}{100}}$ = $\dfrac{log_{10}{2}}{log_{10}{100}}$

$\Rightarrow$ $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$

$\Rightarrow$ $\dfrac{log_{10}{2}}{log_{10}{100}}$-$\dfrac{log_{10}{4}}{log_{10}{100}}$+$\dfrac{log_{10}{5}}{log_{10}{100}}$-$\dfrac{log_{10}{10}}{log_{10}{100}}$+$\dfrac{log_{10}{20}}{log_{10}{100}}$-$\dfrac{log_{10}{25}}{log_{10}{100}}$+$\dfrac{log_{10}{50}}{log_{10}{100}}$

We know that $log_{10}{100}=2$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{\dfrac{2*5*20*50}{4*10*25}}]$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}10]$

$\Rightarrow$ $\dfrac{1}{2}$