CAT Logarithms, Surds & Indices Questions (with Notes) PDF

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CAT Logarithms, Surds and Indices Questions
CAT Logarithms, Surds and Indices Questions

Logarithms, Surds & Indices are important topics in the CAT Quants section. If we look at the Previous year’s papers of CAT Logarithms Surds & Indices have made a recurrent appearance in the CAT Quant Section. It is an important topic and hence must not be avoided by the aspirants. In this article, we will look into some important CAT Logarithms, Surds & Indices Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download these CAT Logarithms Surds & Indices Questions PDF, which is completely Free.

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  • CAT Logarithms Surds & Indices – Tip 1: Questions based on this concept appear in the CAT and other MBA entrance exams every year. If you’re starting the prep, firstly understand the CAT Quants Syllabus;  Based on our analysis of CAT Logarithms Surds & Indices questions in CAT Previous Years this was the weightage: 1-2 questions were asked from these topics (in CAT 2021).
  • CAT Logarithms Surds & Indices – Tip 2: The CAT Logarithms questions are Not very tough to solve. A strong foundation in this topic will help you in answering these questions with ease. Getting yourself acquainted with the basics of these concepts will help you solve the problems. Download this CAT Logarithms questions with solutions PDF (which also includes the Logs and Surds concept PDF). Learn all the major formulae from these concepts (You can learn all the Important CAT Logarithms Surds & Indices Formulas here). Also, do check out the CAT Logarithms surds & indices questions from CAT Mock tests.

Question 1: 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) 5

b) 4

c) 2

d) 3

1) Answer (D)

View Video Solution

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: 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) no solution for x

b) exactly one solution for x

c) exactly two distinct solutions for x

d) exactly three distinct solutions for x

2) Answer (B)

View Video Solution

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: If x = -0.5, then which of the following has the smallest value?

a) $2^{1/x}$

b) $1/x$

c) $1/x^2$

d) $2^x$

e) $1/\sqrt{-x}$

3) Answer (B)

View Video Solution

Solution:

$2^p$ is always positive

$x^2$ is always non negative.

$1/\sqrt{-x}$ is always positive.

$\frac{1}{x}$ is negative when x is negative.

In this case, x is negative => $\frac{1}{x}$ is smallest.

Question 4: Which among $2^{1/2}, 3^{1/3}, 4^{1/4}, 6^{1/6}$, and $12^{1/12}$ is the largest?

a) $2^{1/2}$

b) $3^{1/3}$

c) $4^{1/4}$

d) $6^{1/6}$

e) $12^{1/12}$

4) Answer (B)

View Video Solution

Solution:

Make the power equal and compare the denominators.

$2^{1/2}$ can be written as $64^{1/12}$

$3^{1/3}$ can be written as $81^{1/12}$

$4^{1/4}$ can be written as $64^{1/12}$

$6^{1/6}$ can be written as $36^{1/12}$

Among these, $81^{1/12}$ is the greatest => $3^{1/3}$ is the greatest.

Question 5: 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) (-2, 1/2)

b) (1,1)

c) (0.4, 2.5)

d) ($\pi$, 1/ $\pi$)

e) (2,2)

5) Answer (E)

View Video Solution

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

a) -1

b) -0.5

c) 0

d) 1

6) Answer (D)

View Video Solution

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 7: $2^{73}-2^{72}-2^{71}$ is the same as

a) $2^{69}$

b) $2^{70}$

c) $2^{71}$

d) $2^{72}$

7) Answer (C)

View Video Solution

Solution:

$2^{71} (2^2 – 2^1 – 1)$
$2^{71} (4-2-1)$
$2^{71}$

Question 8: If $\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1, then what could be the value of ‘x’?

a) 3

b) 5

c) 4

d) None of these

8) Answer (C)

View Video Solution

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 9: Suppose, $\log_3 x = \log_{12} y = a$, where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\log_6 G$ is equal to

a) $\sqrt{a}$

b) 2a

c) a/2

d) a

9) Answer (D)

View Video Solution

Solution:

We know that $\log_3 x = a$ and $\log_{12} y=a$
Hence, $x = 3^a$ and $y=12^a$
Therefore, the geometric mean of $x$ and $y$ equals $\sqrt{x \times y}$
This equals $\sqrt{3^a \times 12^a} = 6^a$

Hence, $G=6^a$ Or, $\log_6 G = a$

Question 10: The value of $\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$ is equal to

a) 1/3

b) 2/3

c) 5/6

d) 7/6

10) Answer (C)

View Video Solution

Solution:

$\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$

$81 = 3^4$ and $0.008 = \frac{8}{1000} = \frac{2^{3}}{10^{3}} = \frac{1}{5^{3}} = 5^{-3} $

Hence,

$\log_{0.008}\sqrt{5}+ 8 -7 $

$ \log_{5^{-3}}5^{\frac{1}{2}}+ 8 -7 $

$\frac{log 5^{0.5}}{log 5^{-3}} + 1$

$ – \frac{1}{6} + 1$

= $\frac{5}{6}$

Question 11: If $9^{2x-1}-81^{x-1}=1944$, then $x$ is

a) 3

b) 9/4

c) 4/9

d) 1/3

11) Answer (B)

View Video Solution

Solution:

$\frac{81^x}{9} – \frac{81^x}{81} = 1944$

$81^x * [\frac{1}{9}- \frac{1}{81}] = 1944$

$81^x * [\frac{1}{81}] = 243$

$3^{4x} = 3^9$

$x = \frac{9}{4}$

Question 12: If x is a real number such that $\log_{3}5= \log_{5}(2 + x)$, then which of the following is true?

a) 0 < x < 3

b) 23 < x < 30

c) x > 30

d) 3 < x < 23

12) Answer (D)

View Video Solution

Solution:

$1 < \log_{3}5 < 2$
=> $ 1 < \log_{5}(2 + x) < 2 $
=> $ 5 < 2 + x < 25$
=> $ 3 < x < 23$

Question 13: If $9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$, then $x$ is

a) 3/2

b) 2/5

c) 3/4

d) 4/9

13) Answer (A)

View Video Solution

Solution:

It is given that $9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$

Let us try to reduce them to powers of $3$ and $2$
The given equation can be reduced to $3^{2x-1} + 3^{2x-3} = 2^{2x} + 2^{2x-2}$

Hence, $3^{2x-3} \times 10 = 2^{2x-2} \times 5$
Therefore, $3^{2x-3} = 2^{2x-3}$

This is possible only if $2x-3=0$ or $x=3/2$

Question 14: If $log(2^{a}\times3^{b}\times5^{c} )$is the arithmetic mean of $log ( 2^{2}\times3^{3}\times5)$, $log(2^{6}\times3\times5^{7} )$, and $log(2 \times3^{2}\times5^{4} )$, then a equals

14) Answer: 3

View Video Solution

Solution:

$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $

$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $

$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $

$3log(2^{a}\times3^{b}\times5^{c} )$ = $ log ( 2^{9}\times3^{6}\times5^{12}) $
Hence, 3a = 9 or a = 3

Question 15: If x is a positive quantity such that $2^{x}=3^{\log_{5}{2}}$. then x is equal to

a) $\log_{5}{8}$

b) $1+\log_{3}({\frac{5}{3}})$

c) $\log_{5}{9}$

d) $1+\log_{5}({\frac{3}{5}})$

15) Answer (D)

View Video Solution

Solution:

Givne that: $2^{x}=3^{\log_{5}{2}}$

$\Rightarrow$ $2^{x}=2^{\log_{5}{3}}$

$\Rightarrow$ $x=\log_{5}{3}$

$\Rightarrow$ $x=\log_{5}{\dfrac{3*5}{5}}$

$\Rightarrow$ $x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$

$\Rightarrow$ $x=1+\log_{5}{\dfrac{3}{5}}$. Hence, option D is the correct answer.

Question 16: If $\log_{12}{81}=p$, then $3(\dfrac{4-p}{4+p})$ is equal to

a) $\log_{4}{16}$

b) $\log_{6}{16}$

c) $\log_{2}{8}$

d) $\log_{6}{8}$

16) Answer (D)

View Video Solution

Solution:

Given that: $\log_{12}{81}=p$

$\Rightarrow$ $\log_{81}{12}=\dfrac{1}{p}$

$\Rightarrow$ $\log_{3}{3*4}=\dfrac{4}{p}$

$\Rightarrow$ $1+\log_{3}{4}=\dfrac{4}{p}$

Using Componendo and Dividendo,

$\Rightarrow$ $\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\log_{36}{64}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$. Hence, option D is the correct answer.

Question 17: Given that $x^{2018}y^{2017}=\frac{1}{2}$, and $x^{2016}y^{2019}=8$, then value of $x^{2}+y^{3}$ is

a) $\dfrac{31}{4}$

b) $\dfrac{35}{4}$

c) $\dfrac{37}{4}$

d) $\dfrac{33}{4}$

17) Answer (D)

View Video Solution

Solution:

Given that $x^{2018}y^{2017}=\frac{1}{2}$  … (1)

$x^{2016}y^{2019}=8$ … (2)

Equation (2)/ Equation (1)

$\dfrac{y^2}{x^2} = \dfrac{8}{1/2}$

$\dfrac{y}{x} = 4$ or $-4$

Case 1: When $\dfrac{y}{x} = 4$

$x^{2018}(4x)^{2017}=\dfrac{1}{2}$

$x^{2018+2017}(2)^{4034}=\dfrac{1}{2}$

$x^{4035}=\dfrac{1}{(2)^{4035}}$

$x=\dfrac{1}{2}$

Since, $\dfrac{y}{x} = 4$, => y = 2

Therefore, $x^{2}+y^{3}$ = $\dfrac{1}{4}+8$ = $\dfrac{33}{4}$

Case 2: When $\dfrac{y}{x} = -4$

$x^{2018}(-4x)^{2017}=\dfrac{1}{2}$

$x^{2018+2017}(2)^{4034}=\dfrac{-1}{2}$

$x^{4035}=\dfrac{1}{(-2)^{4035}}$

$x=\dfrac{-1}{2}$

Since, $\dfrac{y}{x} = -4$, => y = 2

Therefore, $x^{2}+y^{3}$ = $\dfrac{1}{4}+8$ = $\dfrac{33}{4}$. Hence, option D is the correct answer.

Question 18: If $\log_{2}({5+\log_{3}{a}})=3$ and $\log_{5}({4a+12+\log_{2}{b}})=3$, then a + b is equal to

a) 59

b) 40

c) 32

d) 67

18) Answer (A)

View Video Solution

Solution:

$\log_{2}({5+\log_{3}{a}})=3$
=>$5 + \log_{3}{a}$ = 8
=>$ \log_{3}{a}$ = 3
or $a$ = 27

$\log_{5}({4a+12+\log_{2}{b}})=3$
=>$4a+12+\log_{2}{b}$ = 125
Putting $a$ = 27, we get
$\log_{2}{b}$ = 5
or, $b$ = 32

So, $a + b$ = 27 + 32 = 59
Hence, option A is the correct answer.

Question 19: If N and x are positive integers such that $N^{N}$ = $2^{160}\ and \ N{^2} + 2^{N}\ $ is an integral multiple of $\ 2^{x}$, then the largest possible x is

19) Answer: 10

View Video Solution

Solution:

It is given that $N^{N}$ = $2^{160}$

We can rewrite the equation as $N^{N}$ = $(2^5)^{160/5}$ = $32^{32}$

$\Rightarrow$ N = 32

$N{^2} + 2^{N}$ = $32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$

Hence, we can say that $N{^2} + 2^{N}$ can be divided by $2^{10}$

Therefore, x$_{max}$ = 10

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

a) $\frac{1}{2}$

b) 10

c) 0

d) −4

20) Answer (A)

View Video Solution

Solution:

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

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

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

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

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

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

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

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

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

Question 21: If p$^{3}$ = q$^{4}$ = r$^{5}$ = s$^{6}$, then the value of $log_{s}{(pqr)}$ is equal to

a) $\frac{47}{10}$

b) $\frac{24}{5}$

c) $\frac{16}{5}$

d) $1$

21) Answer (A)

View Video Solution

Solution:

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

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

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

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

Similarly, r = s$^{\frac{6}{5}}$   …(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 22: The real root of the equation $2^{6x} + 2^{3x + 2} – 21 = 0$ is

a) $\log_{2}{9}$

b) $\frac{\log_{2}{3}}{3}$

c) $\log_{2}{27}$

d) $\frac{\log_{2}{7}}{3}$

22) Answer (B)

View Video Solution

Solution:

Let $2^{3x}$ = v

$2^{6x} + 2^{3x + 2} – 21 = 0$

= $v^2+4v-21=0$

=(v+7)(v-3)=0

v=3, -7

$2^{3x}$ = 3 or $2^{3x}$ = -7(This can be negated)

3x=$\log_23$

x=$\log_23$/3

Question 23: If $(5.55)^x = (0.555)^y = 1000$, then the value of $\frac{1}{x} – \frac{1}{y}$ is

a) $\frac{1}{3}$

b) 3

c) 1

d) $\frac{2}{3}$

23) Answer (A)

View Video Solution

Solution:

We have, $(5.55)^x = (0.555)^y = 1000$

Taking log in base 10 on both sides,

x($\log_{10}555$-2) = y($\log_{10}555$-3) = 3

Then, x($\log_{10}555$-2) = 3…..(1)

y($\log_{10}555$-3) = 3 …..(2)

From (1) and (2)

=> $\log_{10}555$=$\ \frac{\ 3}{x}$+2=$\ \frac{\ 3}{y}+3$

=> $\frac{1}{x} – \frac{1}{y}$ = $\frac{1}{3}$

Question 24: If m and n are integers such that $(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$ then m is

a) -20

b) -24

c) -12

d) -16

24) Answer (C)

View Video Solution

Solution:

We have, $(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$

Converting both sides in powers of 2 and 3, we get

$2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$ = $3^n2^{4m}2^{\frac{\ 6}{4}}$

Comparing the power of 2 we get, $\ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\ $

=> 4m=3n+12 …..(1)

Comparing the power of 3 we get, $4+2m=n$

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12

Question 25: 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) 150

b) 25

c) 100

d) 250

25) Answer (A)

View Video Solution

Solution:

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

=> $x^2-y^2=125$……(1)

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

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

=> 2x=3y   => 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 26: If Y is a negative number such that $2^{Y^2({\log_{3}{5})}}=5^{\log_{2}{3}}$, then Y equals to:

a) $\log_{2}(\frac{1}{5})$

b) $\log_{2}(\frac{1}{3})$

c) $-\log_{2}(\frac{1}{5})$

d) $-\log_{2}(\frac{1}{3})$

26) Answer (B)

View Video Solution

Solution:

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

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

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

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

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

Question 27: If $x=(4096)^{7+4\sqrt{3}}$, then which of the following equals to 64?

a) $\frac{x^{7}}{x^{2\sqrt{3}}}$

b) $\frac{x^{\frac{7}{2}}}{x^{\frac{4}{\sqrt{3}}}}$

c) $\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$

d) $\frac{x^{7}}{x^{4\sqrt{3}}}$

27) Answer (C)

View Video Solution

Solution:

$x=2^{12\left(7+4\sqrt{\ 3}\right)}$.

$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$

$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$

$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$ = $2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$ =$2^6$.

Hence C is correct answer.

Question 28: If $\log_{4}{5}=(\log_{4}{y})(\log_{6}{\sqrt{5}})$, then y equals

28) Answer: 36

View Video Solution

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 29: 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) $\frac{A+B-3}{2}$

29) Answer (A)

View Video Solution

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

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

Question 30: If a,b,c are non-zero and $14^a=36^b=84^c$, then $6b(\frac{1}{c}-\frac{1}{a})$ is equal to

30) Answer: 3

View Video Solution

Solution:

Let  $14^a=36^b=84^c$ = k

=> a = $\log_{14}k$ , b = $\log_{36}k$, c=$\log_{84}k$

$6b(\frac{1}{c}-\frac{1}{a})$ = $6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$ = 3

Question 31: $\frac{2\times4\times8\times16}{(\log_{2}{4})^{2}(\log_{4}{8})^{3}(\log_{8}{16})^{4}}$ equals

31) Answer: 24

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Solution:

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

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

Question 32: 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

32) Answer (C)

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Solution:

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.

Question 33: If $5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$, then 100x equals

33) Answer: 99

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

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

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

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

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

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

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

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

Question 34: If $\log_{2}[3+\log_{3} \left\{4+\log_{4}(x-1) \right\}]-2=0$ then 4x equals

34) Answer: 5

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Solution:

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

Question 35: 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) $a>5$

35) Answer (C)

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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 .

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