Algebra - Logarithms - Previous Year CAT/MBA Questions

Answer: Option D

Explanation :

Given $\frac{\log \left(97-56\sqrt{3}\right)}{\log \left(\sqrt{7+4\sqrt{3}}\right)}$ 

= $\frac{\log {\left(7-4\sqrt{3}\right)}^2}{\log {\left(7+4\sqrt{3}\right)}^{1/2}}$

Now, $7+4\sqrt{3}$ = $\frac{1}{7-4\sqrt{3}}$

  ⇒ $\frac{\log {\left(7-4\sqrt{3}\right)}^2}{\log {\left(7+4\sqrt{3}\right)}^{1/2}}$ = $\frac{\log {\left(7-4\sqrt{3}\right)}^2}{\log {\left(7-4\sqrt{3}\right)}^{-1/2}}$ = $\frac{2}{-1/2}$ = -4

Answer: Option B

Explanation :

Since there is no condition imposed on x, y and z, assume x = y = z = 10

Consider log_xyz + 1.

When x = y = z = 10; log_xyz + 1

= log₁₀(10 × 10) + 1

= log₁₀(10)² + 1 = 2 log₁₀(10) + 1

= 2(1) + 1 = 3

Similarly, log_yxz + 1 = log_zxy + 1 = 3

∴ Required value = (1/3) + (1/3) + (1/3) = 1

Hence, option (b).

Answer: Option B

Explanation :

Consider log_e3 = x.

Hence, the series can also be written as

$\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\right)+\left(5x+\frac{{\left(5x\right)}^2}{2!}+\frac{{\left(5x\right)}^3}{3!}+...\right)$

Now, each bracket is a standard expression that can be expressed as a power of e. Hence, the expression becomes:

$(e^x-1)+(e^{5x}-1)=e^{{\log }_{e}3}-1+e^{5{\log }_{e}3}-1$

= 3 – 1 + 3⁵ – 1 = 1 + 243 = 244

Hence, option (b).

Note:

​​​​​​​This is one of the rare questions where pure engineering maths based concepts have been used. In the exam, you are advised to leave such a question if you are not conversant with such series.

Answer: Option D

Explanation :

4 log₇(x – 8) = log₃ 81

∴ 4 log₇(x – 8) = log₃(3⁴) = 4

∴ log₇(x – 8) = 1

∴ x – 8 = 7 i.e. x = 15

Hence, option (d). 

Answer: Option C

Explanation :

a + b = log₂₅5 + log₂₅15 = log₂₅75 = log₂₅25 + log₂₅3

a + b = 1 + log₂₅3

a + b – 1 = log₂₅3
∴ log₂₅27 = 3log₂₅3 = 3(a + b – 1)

Hence, option (c).

Answer: Option D

Explanation :

${\log }_{10}$ x - ${\log }_{10}$ $\sqrt[3]{x}$ = $\frac{6}{{\log }_{10}x}$

∴ ${\log }_{10}$ x - $\frac{1}{3}$ ${\log }_{10}$ x = $\frac{6}{{\log }_{10}x}$

${\log }_{10}$ x = a

∴ a² - $\frac{1}{3}$ a² = 6

∴ a² = 9

∴ a = ±3

As log cannot be negative a = 3

∴ ${\log }_{10}$ x = 3

∴ x = 1000

Hence, option (d).

Answer: Option D

Explanation :

log₁₃ log₂₁ {$\sqrt{x+21}$ + $\sqrt{x}$} = 0

∴ log₂₁ {$\sqrt{x+21}$ + $\sqrt{x}$} = 13° = 1

∴ {$\sqrt{x+21}$ + $\sqrt{x}$} = 21¹ = 21

∴ $\sqrt{x+21}$= 21 - $\sqrt{x}$

Squaring both sides,

x + 21 = 441 + x - 42$\sqrt{x}$

∴ 42$\sqrt{x}$ = 420

∴ x = 100

Hence, option (d).

Answer: Option B

Explanation :

log(3^x – 2) – log3 = log (3^x + 4) – log(3^x – 2)

Let 3x = t

∴ log$\left\{\frac{t-2}{3}\right\}$ = log$\left\{\frac{t+4}{t-2}\right\}$

∴ $\frac{t-2}{3}$ = $\frac{t+4}{t-2}$

∴t² + 4 – 4t = 3t + 12

∴ t² – 7t – 8 = 0

∴ (3x – 8)(3x + 1) = 0

Since 3^x cannot be negative,

∴ 3^x = 8

∴ x = log₃8

Hence, option (b).

Answer: Option B

Explanation :

log₁₀ 3 + log₁₀ (4x + 1) = log₁₀ (x + 1) + 1

∴ log₁₀ 3 + log₁₀ (4x + 1) – log₁₀ (x + 1) = 1

∴ log₁₀ [3(4x + 1)/(x + 1)] = 1

∴ (12x + 3)/(x + 1) = 10

∴ x = 7/2

Hence, option (b).

Answer: Option C

Explanation :

log₄ log₄ 4^{a - b} = 2log₄ $\left(\sqrt{a}-\sqrt{b}\right)$ + 1

∴ log₄ (a - b) log₄ 4 = 2log₄ $\left(\sqrt{a}-\sqrt{b}\right)$ + 1

∴ log₄ (a - b) = 2log₄ $\left(\sqrt{a}-\sqrt{b}\right)$ + 1

∴ log₄ (a - b) = 2log₄ ${\left(\sqrt{a}-\sqrt{b}\right)}^2$ + 1

∴ log₄ $\left[\frac{a-b}{{\left(\sqrt{a}-\sqrt{b}\right)}^2}\right]$ = 1

∴ $\frac{a-b}{a+b-2\sqrt{ab}}$ = 4

∴ a - b = 4a + 4b - 8$\sqrt{ab}$

∴ 8$\sqrt{ab}$ = 3a + 5b

∴ 8 = 3$\sqrt{\frac{a}{b}}$ + 5$\sqrt{\frac{b}{a}}$

Let $\sqrt{\frac{a}{b}}$ = x

∴ 8 = 3x + $\frac{5}{x}$

∴ 8x = 3x² + 5

∴ 3x² - 8x + 5 = 0

∴ 3x² – 3x – 5x + 5 = 0

∴ 3x(x – 1) – 5(x – 1) = 0

∴(3x – 5)(x – 1) = 0

∴ x = $\frac{5}{3}$ or x = 1

But x ≠ 1 as $\sqrt{a}$ ≠ $\sqrt{b}$

∴ x ≠ 5/3

∴ $\sqrt{\frac{a}{b}}$ = $\frac{5}{3}$

Hence, option (c).

Answer: Option D

Explanation :

Let log₅ 2 be rational.

Then log₅ 2 = $\frac{m}{n}$

∴ $5^{\frac{m}{n}}$ = 2

∴ ${\left(5^{m/n}\right)}^n$ = $2^n$

∴ 5^m = 2ⁿ

However, number 2 raised to any positive integer power must be even, but 5 raised to any positive integer power must be odd.

Hence, we have a contradiction.

∴ log₅ 2 is irrational.

Hence, option (d).

Answer: Option D

Explanation :

S = $\sum _{i=0}^{\infty }$$\frac{1}{(2i+1)(2i+2)(2i+3)}$

= $\frac{1}{2}$ $\sum _{i=0}^{\infty }$$\frac{(2i+3)-(2i+1)}{(2i+1)(2i+2)(2i+3)}$

= $\frac{1}{2}$ $\sum _{i=0}^{\infty }$$\left(\frac{1}{(2i+1)(2i+2)}-\frac{1}{(2i+2)(2i+3)}\right)$

= $\frac{1}{2}$$\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}-\frac{2}{2i+2}+\frac{1}{2i+3}\right)$

= $\frac{1}{2}$$\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}+\frac{1}{2i+3}\right)$ - $\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+2}\right)$            ...(i)

Now, $\frac{1}{2}$$\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}+\frac{1}{2i+3}\right)$ = $\frac{1}{2}$$\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{3}+\frac{1}{5}+\frac{1}{5}+\frac{1}{7}+\frac{1}{7}+...\right)$

= $\frac{1}{2}$$\left(1+2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)\right)$

= $\frac{1}{2}$$\left(2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)-1\right)$

= $\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)$ - $\frac{1}{2}$

= -$\frac{1}{2}$ + $\sum _{i=0}^{\infty }$ $\left(\frac{1}{2i+1}\right)$                                         ...(ii)

From (i) and (ii), we get,

S = -$\frac{1}{2}$ + $\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}-\frac{1}{2i+2}\right)$

= -$\frac{1}{2}$$\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...\right)$

= -$\frac{1}{2}$ + $\sum _{i=0}^{\infty }$${\left(-1\right)}^{(i-1)}$ $\frac{1}{i}$                                ...(iii)

Now, we know that,

log_e (1 + x) = x - $\frac{1}{2}$x² + $\frac{1}{3}$x³ - $\frac{1}{4}$x⁴  + ...

= $\sum _{i=0}^{\infty }$(-1)^{n- 1}  $\frac{1}{n}$xⁿ 

Putting x = 1 in the above equation, we get,

log_e 2 = $\sum _{i=0}^{\infty }$${\left(-1\right)}^{(i-1)}$ $\frac{1}{i}$                            ...(iv)

From (iii) and (iv), we get,

S = log_e 2 - $\frac{1}{2}$

Hence, option (d).

Alternatively,

You can also solve this question using options.

$\frac{1}{1.2.3}$ + $\frac{1}{3.4.5}$ + $\frac{1}{5.6.7}$ + ... = $\frac{1}{6}$ + $\frac{1}{60}$ + $\frac{1}{210}$ + ...

= 0.167 + 0.0167 + 0.0047 + …

< 0.2 but always positive

So, value of S will always be less than 0.2.

Now, we will consider all the given options one by one.

Consider option 1.

e² – 1 = (2.718)² – 1 > 0.2

So, option 1 can be eliminated.

Consider option 2.

log_e2 – 1 = 0.693 – 1 < 0

So, option 2 can be eliminated.

Consider option 3.

2log₁₀2 – 1 = 2(0.3010) – 1 < 0

So, option 3 can be eliminated.

So, the correct answer will be option 4.

Hence, option (d).

Answer: Option B

Explanation :

log₂x.log$\frac{x}{64}$2 = log$\frac{x}{16}$2

∴ $\frac{\log x}{\log x}$ × $\frac{\log 2}{\log \left({\displaystyle \frac{x}{64}}\right)}$ = $\frac{\log 2}{\log \left({\displaystyle \frac{x}{16}}\right)}$

∴ $\frac{\log x\times \log \left({\displaystyle \frac{x}{16}}\right)}{\log {\displaystyle \frac{x}{64}}}$ = log 2

∴ (log x) × $\frac{(\log x-\log 16)}{(\log x-\log 64)}$ = log 2

Let log x = t

∴ $\frac{t(t-\log 16)}{t-\log 64}$ = log 2

∴ t² - t log 16 = t log 2 - log 2. log 64

∴ t² - 4t log 2 = t log 2 - 6(log 2)²

∴ t² - 5t log 2 + 6(log 2)² = 0

∴ (t - 2 log 2) (t - 3 log 2) = 0

∴ t = log 4 or t = log 8

∴ x = 4 or x = 8

Hence, option (b).