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# Logarithm Questions for SNAP – Most Expected in Quant

The logarithm is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Logarithm Questions for SNAP. One can also download this Free Logarithm Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Logarithm questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

Question 1:Â If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7/2)$ are in arithmetic progression, then the value of x is equal to

a)Â 5

b)Â 4

c)Â 2

d)Â 3

Solution:

$2 log (2^x – 5) = log 2 + log (2^x – 7/2)$
Let $2^x = t$
=> $(t-5)^2 = 2(t-7/2)$
=> $t^2 + 25 – 10t = 2t – 7$
=> $t^2 – 12t + 32 = 0$
=> t = 8, 4
Therefore, x = 2 or 3, but $2^x$ > 5, so x = 3

Question 2:Â Let $u = ({\log_2 x})^2 – 6 {\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has

a)Â no solution for x

b)Â exactly one solution for x

c)Â exactly two distinct solutions for x

d)Â exactly three distinct solutions for x

Solution:

$x^u = 256$

Taking log to the base 2 on both the sides,

$u * \log_{2}{x} = \log_{2}{256}$

=>$[({\log_2 x})^2 – 6 {\log_2 x} + 12] * \log_{2}{x} = 8$

$(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$

Let $log_2 x = t$

$t^3 – 6t^2 +12t – 8 = 0$

$(t-2)^3 = 0$

Therefore, $log_2 x = 2$

=> $x = 4$ is the only solution

Hence, option B is the correct answer.

Question 3:Â If $log_y x = (a*log_z y) = (b*log_x z) = ab$, then which of the following pairs of values for (a, b) is not possible?

a)Â (-2, 1/2)

b)Â (1,1)

c)Â (0.4, 2.5)

d)Â ($\pi$, 1/ $\pi$)

e)Â (2,2)

Solution:

$log_y x = ab$
$a*log_z y = ab$ => $log_z y = b$
$b*log_x z = ab$ => $log_x z = a$
$log_y x$ = $log_z y * log_x z$ => $log x/log y$ = $log y/log z * log z/log x$
=> $\frac{log x}{log y} = \frac{log y}{log x}$
=> $(log x)^2 = (log y)^2$
=> $log x = log y$ or $log x = -log y$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible

Question 4:Â If x >= y and y > 1, then the value of the expression $log_x (x/y) + log_y (y/x)$ can never be

a)Â -1

b)Â -0.5

c)Â 0

d)Â 1

Solution:

$log_x (x/y) + log_y (y/x)$ = $1 – log_x (y) + 1 – log_y (x)$
= $2 – (log_x y + 1/log_x y)$ <= 0 (Since $log_x y + 1/log_x y$ >= 2)
So, the value of the expression cannot be 1.

Question 5:Â If $f(x) = \log \frac{(1+x)}{(1-x)}$, then f(x) + f(y) is

a)Â $f(x+y)$

b)Â $f{\frac{(x+y)}{(1+xy)}}$

c)Â $(x+y)f{\frac{1}{(1+xy)}}$

d)Â $\frac{f(x)+f(y)}{(1+xy)}$

Solution:

If $f(x) = \log \frac{(1+x)}{(1-x)}$ then $f(y) = \log \frac{(1+y)}{(1-y)}$

Also Log (A*B)= Log A + Log B

f(x)+f(y) = $\log \frac{(1+x)(1+y)}{(1-x)(1-y)}$

=$\log\frac{\left(1+xy\ +x\ +y\right)}{\left(1+xy-x-y\right)}$

Dividing numberator and denominator by (1+xy)

$\log\frac{\frac{\left(1+xy\ +x\ +y\right)}{1+xy}}{\frac{\left(1+xy-x-y\right)}{1+xy}}$

=$\log\frac{\frac{1+xy\ }{1+xy}+\frac{\left(x+y\right)}{1+xy}}{\frac{1+xy\ }{1+xy}-\frac{\left(x+y\right)}{1+xy}}$

=Â $\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$

Hence option B.

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Question 6:Â If $\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1, then what could be the value of â€˜xâ€™?

a)Â 3

b)Â 5

c)Â 4

d)Â None of these

Solution:

$\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1

$\log_{7}{(x^2 – x+37)}$ = $2$

$(x^2 – x+37)$ = $7^{2}$

Given eq. can be reduced to $x^2 – x + 37 = 49$

So x can be either -3 or 4.

Question 7:Â If $\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$. Then x is

a)Â 2

b)Â 4

c)Â 16

d)Â 12

Solution:

$\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$

i.e. $\frac{log{x}}{log{2}} * \frac{log_{2}}{log{x}-log{64}} = \frac{log{2}}{log{x}-log{16}}$

i.e.Â $\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\log{2}$

let t = log x

Therefore,Â Â $\frac{t * (t-log{16})}{t-log{64}}$ = $\log{2}$

$t^2-4*log 2*t = t*log 2-6*(log 2)^2$

I.e.Â $t^2-5*log 2*t-6*(log 2)^2$ = 0

I.e.Â $t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0

i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0

i.e $t=2*log 2$ or $t=3*log 2$

i.e $log x=log 4$ or $log x=log 8$

therefore $x=4$ or $8$

therefore our answer is option ‘B’

Question 8:Â What is the value of $\sqrt{\frac{a}{b}}$, If $\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+1$

a)Â -5/3

b)Â 2

c)Â 5/3

d)Â 1

Solution:

$\sqrt{\frac{a}{b}}$, If $\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+\log_{4}{4}$

i.e. $\log_{4}\log_{4}4^{a-b}=\log_{4}((\sqrt{a}-\sqrt{b})^2)*4$

i.e. $\log_{4}4^{a-b}=((\sqrt{a}-\sqrt{b})^2)*4$

i.e. (a-b)*$\log_{4}4=((\sqrt{a}-\sqrt{b})^2)*4$

i.e. a-b = 4a+4b-8$\sqrt{ab}$

i.e. 3a + 5b – 8$\sqrt{ab}$ = 0

i.e. $3\sqrt\frac{a}{b}^2$ – 8$\sqrt\frac{a}{b}$+5 = 0

putÂ $\sqrt\frac{a}{b}$ = t

therefore 3$t^2$ – 8t + 5 = 0

solving we get t = 1 or t = $\frac{5}{3}$

i.e.Â $\sqrt\frac{a}{b}$ = 1 orÂ $\frac{5}{3}$

but ifÂ $\sqrt\frac{a}{b}$ = 1 then a=b then $\log_{4}(\sqrt{a}-\sqrt{b})$ will become indefinite

ThereforeÂ Â $\sqrt\frac{a}{b}$ =Â $\frac{5}{3}$

Therefore our answer is option ‘C’

Question 9:Â $\log_{5}{2}$ is

a)Â An integer

b)Â A rational number

c)Â A prime number

d)Â An irrational number

Solution:

Let $\log_{5}{2}$ = y

Let us assumeÂ Â $\log_{5}{2}$ is a rational number.

$\log_{5}{2}$ = p/q, where p and q are co primes.

5^(p/q)=2 => 5^p=2^q.

5^p=5*5*5*5*5*5*5………………p times

2^p=2*2*2*2*2*2*2………………q times

No value of p and q can satisfy the equation. Hence y is an irrational number.

Question 10:Â Find the value of x from the following equation:
$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$

a)Â 2/7

b)Â 7/2

c)Â 9/2

d)Â None of the above

Solution:

$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$ can be written as

$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+\log_{10}{10}$

We know thatÂ $\log_{10}{a}+\log_{10}{b}=\log_{10}{ab}$

$\log_{10}{3*(4x+1)}=\log_{10}{(x+1)*10}$

$12x+3=10x+10$

$x=7/2$. Hence, option B is the correct answer.