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

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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.

• 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

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

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) $1/\sqrt{-x}$

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) $12^{1/12}$

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)

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

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) $2^{72}$

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

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

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

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

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

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

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

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) $1+\log_{5}({\frac{3}{5}})$

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) $\log_{6}{8}$

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) $\dfrac{33}{4}$

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

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

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

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) $1$

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) $\frac{\log_{2}{7}}{3}$

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) $\frac{2}{3}$

Solution:

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

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

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

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) $-\log_{2}(\frac{1}{3})$

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) $\frac{x^{7}}{x^{4\sqrt{3}}}$

Solution:

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

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

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

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

Question 28: 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 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) $\frac{A+B-3}{2}$

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

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

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

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

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

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) $a>5$

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 $\left(\log32\ +\log\ 15\right)=4\log a$
=$\log480=\log a^4$
=$a^4\ =480$