Algebra - Logarithms - Previous Year CAT/MBA Questions

Answer: 47

Explanation :

Given, $\left(\frac{4-{\log }_{2}n}{3-{\log }_{4}n}\right)$ < 0

Case 1: 4 – log₂n < 0 and 3 – log₄n > 0
⇒ log₂n > 4 and log₄n < 3
⇒ n > 16 and n < 64
∴ integral values of n can be 17, 18, …, 63 i.e., 47 values.

Case 2: 4 – log₂n > 0 and 3 – log₄n < 0
⇒ log₂n < 4 and log₄n > 3
⇒ n < 16 and n > 64
∴ No integral values of n is possible.

Hence, 47.

Answer: 99

Explanation :

Given,
5 – ${\log }_{10}\left(\sqrt{1+x}\right)$ + 4${\log }_{10}\left(\sqrt{1-x}\right)$ = ${\log }_{10}\left(\frac{1}{\sqrt{1-x^2}}\right)$

⇒  5 – ${\log }_{10}{\left(1+x\right)}^{1/2}$ + ${\log }_{10}{\left(1-x\right)}^{4/2}$ = ${\log }_{10}{\left(1-x^2\right)}^{-1/2}$

⇒  5 = ${\log }_{10}{\left(1+x\right)}^{1/2}$ - ${\log }_{10}{\left(1-x\right)}^2$ - ${\log }_{10}{\left(1-x^2\right)}^{1/2}$

⇒  5 = ${\log }_{10}\left[\frac{{\left(1+x\right)}^{1/2}}{{(1-x)}^2\times {\left\{\right(1+x\left)\right(1-x\left)\right\}}^{1/2}}\right]$

⇒  5 = ${\log }_{10}{(1-x)}^{-5/2}$

⇒ 5 = -5/2 × log₁₀(1 - x)

⇒ -2 = log₁₀(1 - x)

⇒ 1 – x = 10^-2 = 1/100

⇒ 100 – 100x = 1

⇒ 100x = 99

Hence, 99.

Answer: 5

Explanation :

Given, log₂⁡[3 + log₃{4 + log₄(x - 1)}] - 2 = 0

⇒ log₂⁡[3 + log₃{4 + log₄(x - 1)}] = 2

⇒ 3 + log₃{4 + log₄(x - 1)} = 4

⇒ log₃{4 + log₄(x - 1)} = 1

⇒ 4 + log₄(x - 1) = 3

⇒ log₄(x - 1) = -1

⇒ x – 1 = ¼

⇒ x = 5/4

⇒ 4x = 5

Hence, 5.

Answer: Option A

Explanation :

$\frac{{\log }_{15}a+{\log }_{32}a}{{\log }_{15}a\times {\log }_{32}a}$ = 4

Converting all logs to base 10.

$\frac{{\displaystyle \frac{1}{{\log }_{a}15}}+{\displaystyle \frac{1}{{\log }_{a}32}}}{\frac{1}{{\log }_{a}15}\times \frac{1}{{\log }_{a}32}}$ = 4

⇒ log_a32 + log_a15 = 4

⇒ log_a(32 × 15) = 4

⇒ a⁴ = 480

This is possible when 4 < a < 5.

Hence, option (a).

Answer: Option D

Explanation :

Given, $2^{{\mathrm{y}}^2{\log }_{3}5}=5^{{\log }_{2}3}$, 

Taking log on both sides (Choosing the base to be 3)

⇒ y² × log₃5 × log₃2 = log₂3 × log₃5

⇒ y² × log₃2 = log₂3

⇒ y² = (log23)²

⇒ y = - log₂3 (∵ y is a negative number)

⇒ y = log₂(1/3)

Hence, option (d).

Answer: 36

Explanation :

Given, log₄5 = log₄y × log₆√5

⇒ log₄5 × log_y4 = log₆√5

⇒ log_y5 = ½ × log₆5

⇒ log_y5 = log₃₆5

⇒ y = 36.

Hence, 36.

Answer: Option C

Explanation :

${\log }_a\left(\frac{a}{b}\right)+{\log }_b\left(\frac{b}{a}\right)$ = 1 - log_ab + 1 - log_ba = 2 – (log_ab + log_ba)

We know logab is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e.,

∴ log_ab + log_ba ≥ 2

⇒ 2 – (log_ab + log_ba) ≤ 2 – 2

⇒ 2 – (log_ab + log_ba) ≤ 0

∴ 2 – (log_ab + log_ba) cannot take positive value i.e., it cannot be equal to 1 (option (b))

Hence, option (c).

Answer: 24

Explanation :

Given, $\frac{2\times 4\times 8\times 16}{\left({{\log }_{2}4)}^2\right({{\log }_{4}8)}^3({{\log }_{8}16)}^4}$

= $\frac{2\times 2^2\times 2^3\times 2^4}{\left({{\log }_2{2}^2)}^2\right({{\log }_{2^2}{2}^3)}^3({{\log }_{2^3}{2}^4)}^4}$

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

= $\frac{2^{10}\times 2^3\times 3^4}{2^2\times 3^3\times 2^8}$

= 2³ × 3 = 24 

Hence, 24.

Answer: Option C

Explanation :

Since all the 4 options have A + B, lets add A and B.

∴ A + B = log_a30 + (-log_a(5/3)) 

= log_a30 – log_a(5/3) = log_a(30 × 3/5) 

= log_a(18) = log_a(2 × 3²)

⇒ A + B = log_a(2) + 2log_a(3)

⇒ A + B = 3 + 2log_a(3) [∵ log₂a = 1/3 ∴ log_a2 = 3]

⇒ log_a(3) = (A + B - 3)/2

⇒ log₃(a) = 2/(A + B - 3)

Hence, option (c).

Answer: Option B

Explanation :

log₅(x + y) + log₅(x - y) = 3

∴ log₅(x + y)(x - y) = 3.

∴ (x + y)(x − y) = 5³ = 125

So, x² - y² = 125  ....(1)

log₂y - log₂x = 1 - log₂3

∴ ${\log }_2\left(\frac{3y}{x}\right)$ = 1

∴ (3y/x) = 2¹ = 2.

So, 3y = 2x .... (2)

Solving (1) and (2), we get;

x = 15 and y =10.

∴ xy = 15 × 10 = 150. 

Hence, option (b).

Answer: Option A

Explanation :

Given: 2^6x + 2^3x+2 - 21 = 0

Replace 2^3x with y

So, 2^6x = y²

Now, 2^6x + 2^3x+2 - 21 = 0 can be rewritten as y² + 2² × 2^3x - 21 = 0

y² + 4y - 21 = 0

Solving the above quadratic equation,

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

So, y = -7 or +3

y = - 7 is rejected (since, y = 2^3x which should always be positive)

⇒ 2^3x = 3

Taking log on both sides,

log₂⁡3 = 3x 

x = $\frac{{\log }_2\left(3\right)}{3}$ 

Hence, option (a).

Answer: Option D

Explanation :

For roots of a quadratic equation to be real and distinct, Discriminant > 0

So, for x² − 4x – log₂A = 0,

D = (-4)² - (4 × 1 × (-log₂A)) > 0

⇒ 16 + 4 × log₂A > 0

⇒ log₂A > -4

⇒ A > 2^-4

⇒ A > 1/16

Hence, option (c).

Answer: Option C

Explanation :

It is given that, $\sqrt{{\log }_e\left(\frac{4x-x^2}{3}\right)}$ is a real number

Therefore, ${\log }_e\left(\frac{4x-x^2}{3}\right)$ ≥ 0

⇒ $\frac{4x-x^2}{3}$ ≥ 1

⇒ 4x – x² ≥ 3

⇒ x² - 4x + 3 ≤ 0

⇒ (x - 1)(x - 3) ≤ 0

⇒ x ∈ [1, 3]

Hence, option (c).

Answer: Option D

Explanation :

Given, 2^x = $3^{{\log }_5\left(2\right)}$

⇒ 2^x = $2^{{\log }_5\left(3\right)}$

⇒ x = log₅3

Option (a): 1 + ${\log }_3\left(\frac{5}{3}\right)$ = 1 + log₃5 - log₃3 = log₃5

Option (b): log₅9 = 2(log₅3)

Option (c): log₅8 = 3(log₅2)

Option (d): 1 + ${\log }_5\left(\frac{3}{5}\right)$ = 1 + log₅3 - log₅5 = log₅3

Hence, option (d).

Answer: Option C

Explanation :

log₁₂81 = p

⇒ p = 4log₁₂3

Now, $\frac{4-p}{4+p}=\frac{4-4{\log }_{12}3}{4+4{\log }_{12}3}$

= $\frac{1-{\log }_{12}3}{1+{\log }_{12}3}$

= $\frac{{\log }_{12}12-{\log }_{12}3}{{\log }_{12}12+{\log }_{12}3}=\frac{{\log }_{12}12/3}{{\log }_{12}12\times 3}=\frac{{\log }_{12}4}{{\log }_{12}36}$

= $\frac{2{\log }_{12}2}{2{\log }_{12}6}$ = ${\log }_{6}2$

Hence, $3\left(\frac{4-p}{4+p}\right)=3{\log }_{6}2={\log }_{6}8$

Alternately,
log₁₂81 = p
⇒ 81 = 12^p
⇒ 3⁴ = 3^p × 2^2p 
​​​​​​​⇒ 3^{(4 − p)} = 2^2p

Taking log on both the sides,

(4 – p) (log 3) = (2p) (log 2)

∴ $\frac{\log 3}{\log 2}$ = $\frac{2p}{4-p}$

∴ $\frac{\log 3+\log 2}{\log 2}$ = $\frac{2p + (4-p)}{4-p}$ …(by adding 1 both sides)

∴ $\frac{\log 6}{\log 2}$ = $\frac{4+p}{4-p}$

∴ $\frac{4-p}{4+p}$ = $\frac{\log 2}{\log 6}$ = log₆2

∴  $3\left(\frac{4-p}{4+p}\right)$ = 3log₆2 = log₆8

Hence, option (c).

Answer: Option C

Explanation :

log₂(5 + log₃a) = 3
⇒ (5 + log₃a) = 2³ = 8
⇒ log₃a = 3
⇒ a = 3³ = 27

log₅(4a + 12 + log₂b) = 3
⇒ log₅(4 × 27 + 12 + log₂b) = 3
⇒ 120 + log₂b = 5³
⇒ log₂b = 125 - 120
⇒ log₂b = 5
⇒ b = 2⁵ = 32

∴ a + b = 27 + 32 = 59

Hence, option (c).

Answer: Option A

Explanation :

Given, p³ = q⁴ = r⁵ = s⁶ = k

⇒ p³ = k
⇒ p = k^1/3

Similarly,
q = k^1/4
r = k^1/5
s = k^1/6

pqr = k^1/3 × k^1/4 × k^1/5 = k^{1/3 + 1/4 + 1/5} = k^47/60

Now, log_s(pqr) = log_k^1/6 k^47/60 = 47/10

Hence, option (a).

Answer: Option C

Explanation :

We have, 4ⁿ > 17¹⁹

Taking log on both sides to the base 4,
 
n × log₄ 4 > 19 × log₄ 17

Now, 17 > 4^2.

∴ log₄ 17 > 2.

⇒ n × log₄ 4 > 19 × log₄ 17 > 19 × 2

Also, log₄ 4 = 1

∴ n > 38.

Hence, option (c).

Answer: Option D

Explanation :

$\frac{1}{l{\mathrm{og}}_{2}100}$ - $\frac{1}{{\log }_{4}100}$ + $\frac{1}{{\log }_{5}100}$ - $\frac{1}{{\log }_{10}100}$ + $\frac{1}{{\log }_{20}100}$ - $\frac{1}{{\log }_{25}100}$ + $\frac{1}{{\log }_{50}100}$

= ${\log }_{100}2$ - ${\log }_{100}4$ + ${\log }_{100}5$ - ${\log }_{100}10$ + ${\log }_{100}20$ - ${\log }_{100}25$ + ${\log }_{100}50$

= ${\log }_{100}\left(\frac{2\times 5\times 20\times 50}{4\times 10\times 25}\right)$ = ${\log }_{100}\left(10\right)$ = $\frac{1}{2}$

Hence, option (d).

Answer: Option D

Explanation :

log₃x = a ⇒ x = 3^a.

log₁₂y = a ⇒ y = 12^a.

∴ xy = 36^a and √(xy) = G = 6^a.

∴ log₆G = a.

Hence, option (d).

Answer: Option C

Explanation :

0.008 = 8/1000 = 1/125 = 5^-3

∴ log_0.008√5 = ${\log }_{5^{-3}}\left(5^{1/2}\right)$ = $\frac{1/2}{-3}$ = $\frac{-1}{6}$ and

log_√381 = ${\log }_{3^{1/2}}\left(3^4\right)$ = $\frac{4}{1/2}$ = 8.

∴ The given expression is -1/6 + 8 - 7 = 5/6

Hence, option (c).

Answer: Option D

Explanation :

Given log₃5 = log₅(x + 2)

Now log₃3 < log₃5 < log₃9

∴ 1 < log₃5 < 2
⇒ 1 < log₅(x + 2) < 2
⇒ log₅5 < log₅(x + 2) < 2log₅5
⇒ log₅5 < log₅(x + 2) < log₅25
⇒ 5 < 2 + x < 25
⇒ 3 < x < 23

Hence, option (d).

Answer: 3

Explanation :

Arithmetic mean of log(2² × 3³ × 5), (log 2⁶ × 3 × 5⁷) and (log 2 × 3² × 5⁴)

= $\frac{1}{3}$[log(2² × 3³ × 5) + log(2⁶ × 3 × 5⁷) + log(log 2 × 3² × 5⁴)]

= ​​​​​​​$\frac{1}{3}$[log(2² × 3³ × 5 × 2⁶ × 3 × 5⁷ × 2 × 3² × 5⁴)]

= $\frac{1}{3}$[log(2^{2 + 6 + 1}) × (3^{3 + 1 + 2}) × (5^{1 + 7 + 4})]

= ${\log }_{}\left[2^{9\times \frac{1}{3}}\times 3^{6\times \frac{1}{3}}\times 5^{12\times \frac{1}{3}}\right]$

= log [2³ × 3² × 5⁴]

Now log(2^a × 3^b × 5^c) = log(2³ × 3² × 5⁴)

(Note: In the given question it should have been specified that a, b, c are integers, as there can be multiple answers for the same. Since it is a TITA question, we have to assume the same and equate power of 2, 3 and 5)

∴ Equating powers of 2, 3 and 5 we get

a = 3, b = 2 and c = 4

Hence, 3.

Answer: Option E

Explanation :

If log_yx = a × log_zy = b × log_xz = ab

⇒ log_yx = ab   ...(1)

⇒ a × log_zy = ab ⇒ log_zy = b   ...(2)

⇒ b × log_xz = ab ⇒ log_xz = a   ...(3)

From (1), (2) and (3), we get
log_yx = log_xz × log_zy

∴  $\frac{\log x}{\log y}$ = $\frac{\log z}{\log x}\times \frac{\log y}{\log z}$ = $\frac{\log y}{\log x}$

∴ (log x)² = (log y)²

∴ log x = ± log y

∴ log x = log y or log x = - log y

∴  x = y or x = $\frac{1}{y}$

∴ ab = log_y x = 1 or -1

Only option (e) does not satisfy this.

Hence, option (e).

Answer: Option D

Explanation :

${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = log_x x - log_x y + log_y y - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 1 - log_x y + 1 - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - log_x y - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - (log_x y + log_y x)

As x ≥ y and y > 1,
log_y x ≥ 0
Now, (log_x y + log_y x) = ${\log }_x\left(y\right)+\frac{1}{{\log }_x\left(y\right)}$ [This is sum of a positive number and its reciprocal]

Now, sum of a positive number and its reciprocal is always greater than or equal to 2.

∴ (log_x y + log_y x) = ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ ≥ 2

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - (log_x y + log_y x) ≤ 0

∴ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ ≠ 1

Hence, option (d).