Logarithm Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Logarithm topic of quantitative aptitude
(a) –4
(b) 4
(c) 3
(d) –3
The correct answers to the above question in:
Answer: (a)
Let $log_{10}$ 0.0001 = a
a = $log_{10} {1/{(10)^4}$
= $log_{10} 1 – log_{10} (10)^4$ = 0 – 4 = –4
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
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 »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 : 2
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 »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 : 3
$\text"log" 216 √6$ to the base 6 is
a) 7/2
b) 3
c) 2/7
d) 3/2
e) None of these
Answer »Answer: (a)
$\text"log"_{6} 216 √6 = \text"log"_6 (6)^3 (6)^{1/2}$
= $\text"log"_6 (6)^{7/2} = {7/2}{ \text"log"_6} 6 = 7/2$ (∵ $\text"log"_a$ a = 1)
Question : 4
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
Answer »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
Question : 5
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 »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 : 6
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 »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, ${\text"log"_b}(ab) = {\text"log" ab}/{\text"log" b} = {\text"log" a + \text"log" b}/{\text"log" b}$
= ${\text"log" a}/{\text"log" b} + 1 = {1/{x - 1}} + 1 = {1 + x - 1}/{x - 1} = {x}/{x - 1}$
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