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) $log_{10}5$
(b) 0
(c) 1
(d) None of these
The correct answers to the above question in:
Answer: (a)
$log_{10}(3/2) + log_{10}(4/3) + log_{10}(5/4)+ ..... + 8^{th} term$
= $log_{10}(3/2) + log_{10}(4/3)+ log_{10}(5/4)+ ..... +log_{10}({10}/9)$
= $log_{10}(3/2 × 4/3 × 5/4 ×......× {10}/9)$
= $log_{10}({10}/2)log_{10} 5$.
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Read more model question set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
If ${\text"log"_8} x + {\text"log"_8} {1/6} = 1/3$, then find the value of x.
a) 16
b) 18
c) 12
d) 24
e) None of these
Answer »Answer: (c)
${\text"log"_8}x +{ \text"log"_8} {1/6} = 1/3$
or ${\text"log"_8} (x × {1/6}) = 1/3 or {\text"log"_8} (x/6) = 1/3$
or $x/6 = (8)^{1/3}$ {∵ $\text"log"_a$ b = x⇔(a)x=b}
or ${x/6} = (2^3)^{1/3}$ or x = 12
Question : 2
It is given that $log_{10} 2$ = 0.301 and $log_{10} 3 = 0.477.$ How many digits are there in $(108)^{10}$ ?
a) 21
b) 19
c) 20
d) 22
Answer »Answer: (a)
$log (108)^{10} = 10 log 108 = 10 log (2^2 × 3^3)$ = 10 (2log2 + 3log3)
= 10 (2 × 0.301 + 3 × 0.477 ) = 10 (.602 + 1.431)
= 10 × 2.033 = 20.33
integral part = 20
No. of digits = 20 + 1 = 21
Question : 3
What is the solution of $log_{10} [1- [1 - (1 - x^2)^{-1}]^{-1}]^{- 1/2}$ = 1 ?
a) x = 1
b) x = 100
c) x = 10
d) x = 0
Answer »Answer: (c)
$log_{10} [1 - [1 - (1 - x^2)^{-1}]^{-1}]^{-1/2}$ = 1
⇒$log_{10}[1 - [1 - 1/{1 - x^2}]^{-1}]^{-1/2}$ = 1
⇒$log_{10}[1 - [{- x^2}/{1 - x^2}]^{-1}]^{- 1/2}$ = 1
⇒$log_{10}[1 - {(1 - x^2)}/{- x^2}]^{-1/2}$ = 1
⇒$log_{10}[{-x^2 - (1 - x^2)}/{- x^2}]^{-1/2}$ = 1
⇒$log_{10}[1/{x^2}]^{-1/2}$ = 1
⇒$log_{10} x = log_{10}10$
x = 10
⇒$log_{10}$
Question : 4
The sides of a triangle are 30 cm, 28 cm and 16 cm respectively. In order to determine its area, the logarithm of which of the quantities are required ?
a) 37, 21, 11, 9
b) 37, 11, 28, 16
c) 21, 30, 28, 7
d) 37, 21, 9, 7
Answer »Answer: (d)
Semiperimeter of triangle = ${30 + 16 + 28}/2$ = 37
P – a = 7
P – b = 9
P – c = 21
P = 37
∴ quantities that are required
= 37, 21, 9, 7
Question : 5
What is the solution of the equation x $log_{10} ({10}/3) + log_{10} 3 = log_{10} (2 + 3^x)$ + x ?
a) 1
b) 10
c) 3
d) 0
Answer »Answer: (d)
$x log_{10}({10}/3) + log_{10} 3 = log_{10} (2 + 3^x) + x$
$xlog_{10} 10 - x log_{10} 3 + log_{10} 3 = log_{10} (2 + 3^x)$ + x
x - $log_{10^3^x} + log_{10^3} = log_{10} (2 + 3^x)$ + x
$log_{10[3/{3^x}]} log_{10} (2 + 3^x)$
⇒$3^{1–x} = 2 + 3^x$
⇒$3^{1–x} –3^x = 3^1 – 3^0$
x = 0
Question : 6
If ${\text"log"_a} b = 1/2, {\text"log"_b}$ c = $1/3$ and ${\text"log"_c}$ a = $k/5$, then the value of k is
a) 30
b) 25
c) 20
d) 35
e) None of these
Answer »Answer: (a)
${\text"log"_a} b = {1/2}, {\text"log"_b}c= {1/3}$, and ${\text"log"_c} a = k/5$
⇒${\text"log" b}/{\text"log" a} = {1/2}, {\text"log" c}/{\text"log" b} = {1/3}, {\text"log" a}/{\text"log" c} = k/5$
⇒${1/2} × {1/3} × {k/5}$ = 1⇒k = 30
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