Logarithm Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 1 EXERCISES

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The following question based on Logarithm topic of quantitative aptitude

Questions : Let XYZ be an equilateral triangle in which XY = 7 cm. If A denotes the area of the triangle, then what is the value of $log_{10}A^4$ ? (Given that $log_{10}1050 = 3.0212$ and $log_{10}35$ = 1.5441)

(a) 5. 5635

(b) 5. 3070

(c) 5. 3700

(d) 5. 6535

The correct answers to the above question in:

Answer: (b)

$log_{10} 1050 = log_{10}$(3 × 10 × 35)

= $log_{10} 3 + 1 + log_{10} 35$

⇒ 3.0212 = $log_{10} 3 + 1 + 1.5441$

⇒$log_{10} 3$ = 0.4771

Now, $log_{10} 35 = log {7 × 10}/2$

= $log_{10} 7 + log_{10} - log_{10} 2$

⇒ $log_{10}7$ = 0.8451

Now, A = ${√3}/4 × (7)^2$

$log{10^A^4} = 4 log_{10}A$

= 4 log ${√3 × 7^2}/4$

= $4[1/2 log_3 + 2 log_{10} 7 - 2 log 2]$ = 5.3070

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Read more model question set 1 Based Quantitative Aptitude Questions and Answers

Question : 1

What is the logarithm of 0.0001 with respect to base 10?

a) –4

b) 4

c) 3

d) –3

Answer: (a)

Let $log_{10}$ 0.0001 = a

a = $log_{10} {1/{(10)^4}$

= $log_{10} 1 – log_{10} (10)^4$ = 0 – 4 = –4

Question : 2

What is the number of digits in $2^{40}$ ? (Given that $log_{10} 2$ = 0.301)

a) 12

b) 14

c) 13

d) 11

Answer: (c)

let $2^{40} = 10^x$

log $2^{40} = log 10^x$

40 $log^2 = x log^{10}$

40 (.301) = x

12.04 = x

i.e. x > 12

∴ x = 13

No. of digit in $2^{40}$ is 13

Question : 3

If ${\text"log"_7}{\text"log"_5} (√x + 5 + √x)$ = 0 , find the value of x.

a) 2

b) 1

c) – 2

d) 0

e) None of these

Answer: (d)

$\text"log"_7 \text"log"_5(√x + 5 + √x)$ = 0

use $\text"log"_a$ x = b⇒$a^b$=x

∴ $\text"log"_5 (√x + 5 + √x) = 7^0$ = 1

$√x + 5 + √ x = 5^1 = 5⇒2√x$ = 0

∴ x = 0

Question : 4

If $log_r 6 = m and log_r 3$ = n, then what is $log_r (r/2)$ equal to?

a) 1 – m – n

b) m – n + 1

c) m + n – 1

d) 1 – m + n

Answer: (d)

Given, $log_r 6 = \text"m and" log_r$ 3 = n

∵ $log_r 6 = log_r$ (2 × 3)

= $log_r 2 + log_r$ 3

∴ $log_r 3 + log_r$ 2 = m

⇒n + $log_r2$ = m

⇒$log_r$ 2 = m – n

∴ $log_r (r/2) = log_r r - log_r 2$

= 1 – m + n

Question : 5

If ${\text"log"_a}$ (ab ) = , then $\text"log"_b$ (ab) is

a) $x/{x - 1}$

b) $1/x$

c) $x/{1 - x}$

d) $x/{x + 1}$

e) None of these

Answer: (a)

${\text"log"_a}$(ab) = x

⇒${\text"log" ab}/{\text"log" a} = x⇒{\text"log" b}/{\text"log" a} = x - 1⇒{\text"log" a}/{\text"log" b} = 1/{x - 1}$

Now, ${\text"log"_b}(ab) = {\text"log" ab}/{\text"log" b} = {\text"log" a + \text"log" b}/{\text"log" b}$

= ${\text"log" a}/{\text"log" b} + 1 = {1/{x - 1}} + 1 = {1 + x - 1}/{x - 1} = {x}/{x - 1}$

Question : 6

What is the value of $(log_{1/2}2)(log_{1/3}3)(log_{1/4}4).......(log_{1/{1000}}1000)$ ?

a) 1 or –1

b) 1

c) –1

d) 0

Answer: (c)

$(log_{1/2}2)(log_{1/3}3)(log_{1/4}4).......(log_{1/{1000}}1000)$

= $({log 2}/{log_{1/2}})({log 3}/{log_{1/3}})({log 4}/{log_{1/4}})......({log 1000}/{log_(1/{1000})})$

$(∵ log_b a = {log a}/{log b})$

= $({log 2}/{- log 2})({log 3}/{- log 3})({log 4}/{- log 4}).........({log 1000}/{- log 1000})$

= (-1) × (-1) × (-1) × ..... × (-1)

(∵ number of factors is odd)

= –1

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Logarithm Model Questions Set 1

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